Integral Calculus Notes | EduRev

IIT JAM Mathematics

IIT JAM : Integral Calculus Notes | EduRev

The document Integral Calculus Notes | EduRev is a part of the IIT JAM Course IIT JAM Mathematics.
All you need of IIT JAM at this link: IIT JAM

Differentiation Under the Integral Sign

In this section, we shall prove that, under suitable conditions, ‘the derivative of the integral and the integral of the derivative are equal’, and consequently, ‘the two repeated integrals are equal for continuous functions.’

Theorem (Leibnitz’s Rule). If f is defined and continuous on the rectangle R = [a, b; c d], and if

(i) fx(x, y) exists and is continuous on the rectangle R, and

(ii) g(x) = Integral Calculus Notes | EduRev

then g is differentiable on [a, b], and

Integral Calculus Notes | EduRev

i.e., Integral Calculus Notes | EduRev

Corollary 1. (General Leibnitz’s rule). If f satisfy the conditions of the above theorem, and if

(i) θ, ψ : [a, b] → [c, d] are both differentiable, and

(ii) Integral Calculus Notes | EduRev

then g is differentiable on [a, b] and

Integral Calculus Notes | EduRev

Corollary 2. If f is continuous on R = [a, b; c, d], then

Integral Calculus Notes | EduRev

Iterated Integrals (or repeated integrals)

Definition. An iterated integral is an integral of the form

Integral Calculus Notes | EduRev

where θ1or θ2 or both are functions of x or constants.

This means that for each fixed x between a and b, the integral

Integral Calculus Notes | EduRev

is evaluated, and then the integral Integral Calculus Notes | EduRev

or Integral Calculus Notes | EduRev  ----(1)

The other repeated integral

Integral Calculus Notes | EduRev ---(2)

is defined in the same way.

Example 1: If I a I < 1. show that

Integral Calculus Notes | EduRev

We write

Integral Calculus Notes | EduRev

It may be seen that the integrand has only a removable discontinuity at π/2 as much as

Integral Calculus Notes | EduRev

Also Integral Calculus Notes | EduRev

We see that f and fa are both continuous functions of two variables in the domain {(a,x) : 0 ≤ x ≤ π, I a I < 1}. If we assign to f, the value a for x = π/2.

We write

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev   ----(1)

Put Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

put tan (x/2) = t

diff w. r. to x we get

sec2(x/2).1/2 dx = dt

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Therefore, Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

From this we obtain 

∅(a) = πsin-1 a + c. . ..( 2 ) 

where, c, is an arbitrary constant.

From (1), we have 

∅(0) = 0.

Putting a = 0 in (2), we obtain c = 0. 

Hence ∅(a) = πsin-1 a.

Fundamental Theorem of Calculus

Theorem : If a function f is bounded and integrable on [a, b], then the function F defined as

Integral Calculus Notes | EduRev

is continuous on [a, b], and furthermore, if f is continuous at a point c of [a, b], then F is derivable at c and

F’(c) = f(c).

Since f is bounded therefore ∃ a number K > 0 such that 

I f(x) I < K, ∀ x ∈ [a, b].

If x1, x2 are two points of [a, b] such that a ≤ x1 < x2 ≤ b, then

Integral Calculus Notes | EduRev

Thus for a given ∈ > 0, we see that

Integral Calculus Notes | EduRev

Hence the function F is continuous (in fact uniformly) on [a, b].

Let f be continuous at a point c of [a, b], so that for any ∈ > 0 there exists δ > 0 such that 

Integral Calculus Notes | EduRev

Let c - δ < δ < c < t < c + δ

Integral Calculus Notes | EduRev

⇒ F'(c) = f(c).

i.e., continuity of f at any point of [a, b] implies derivability of F at that point.

Note. As c is any point of [a, b], we have for all x ∈ [a, b],

F’(t) = f(t) ⇒ F = f

i.e.; continuity of f on [a, b] implies derivability of F on [a, b].

This theorem is sometimes referred to as the First Fundamental Theorem of Integral Calculus.

Definition : A derivable function F, if it exists such that its derivative F’ is equal to a given function f, is called a primitive of f.

The above theorem shows that a sufficient condition for a function to admit of a primitive is that it is continuous. Thus every continuous function f possesses a primitive F where

Integral Calculus Notes | EduRev

Remark : We shall now show, with the help of an example, that continuity of a function is not a necessary condition for the existence of a primitive, in other words, “functions possessing primitives are not necessarily continuous”.

Consider the function f on [0, 1], where

Integral Calculus Notes | EduRev

It has a primitive F. where

Integral Calculus Notes | EduRev

Clearly F’(x) = f(x) but f(x) is not continuous at x = 0, i.e., f(x) is not continuous on [0, 1] In fact, all this amounts to saying that the derivative of a function is not necessarily continuous. 

Theorem : Function f is bounded and integrable on [a, b] and there exists a function F such that F’ = f on [a, b], then

Integral Calculus Notes | EduRev

Since the function F'. f is bounded and integrable, therefore for every given ∋ > 0, ∃ δ> 0 such that for every partition P = { a = x0 x1  ..x2..... xn = b}. with norm μ(p) < δ,

or Integral Calculus Notes | EduRev ----(1)

for every choice of points t1 in Δx1..

Since we have freedom in the selection of points tt Δxt, we choose them in a particular way as follows:

By Lagrange’s mean value theorem, we have

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= F(b) - F(a).

Hence from (1).

Integral Calculus Notes | EduRev

It is sometimes referred to as The second Fundamental Theorem of Integral Calculus.

Example 2 : Show that the function [x], where [x] denotes the greatest integer not greater than x, is integrable in [0, 3].

Since the function is bounded and has only three points of discontinuity therefore it is integrable and

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

1, 2 and 3 being respectively the points of discontinuity of the three integrals on the right.

Example 3 : f is a non-negative continuous function on [a, b] and Integral Calculus Notes | EduRev. Prove that f(x) = 0, for all x e∈ [a, b].

Suppose that, for some c ∈ (a, b), f(c) > 0

Then, for Integral Calculus Notes | EduRev continuity of f at c implies that, there exists a δ > 0 such that

Integral Calculus Notes | EduRev

Now Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

which is a contradiction. Thus f(x) = 0, ∀ x ∈ (a, b). 

Similarly, f(a) > 0, and f(b) > 0. Hence the result follows.

Double Integral

The Calculation of a double integral 

Equivalence of a double with repeated integrals.

Theorem. If the double integral

Integral Calculus Notes | EduRev

exists where R is the rectangle [a, b; c, d]

and if also Integral Calculus Notes | EduRev exist ∀ y ∈ [c, d], 

then the repeated integral

Integral Calculus Notes | EduRev

exists and is equal to the double integral.

Definite Double Integral : If we have function f(x, y) then

Integral Calculus Notes | EduRev

If dx first then function of y is constant and if dy first then function of x is constant.

Cor. If a double integral exists, then the two repeated integrals cannot exist without being equal. 

Cor. A function f is defined in [0, 1; 0, 1] as-follows :

f(x, y) = 1/2, when y is rational,

f(x, y) = x, when y is irrational;

Integral Calculus Notes | EduRev exists and is equal to 1/2, but the double integral and the second repeated integral do not exist.

Example 4 : Evaluate Integral Calculus Notes | EduRev over the area between y = x2 and y = x.

The area is bounded by the curves 

y = f1(x) = x2, y = f2(x) = x. see in fig.

Integral Calculus Notes | EduRev

When f1(x) = f2(x), x2 = x, i.e., x = 0, x = 1, i.e., the area of integration is bounded by 

y = x2, y = x, x = 0 , x = 1 

In fig. x varies from x = 0 to x = 1 and y varies from y = x2 to y = x

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= 3/56

Example 5 : Prove that

Integral Calculus Notes | EduRev

LHS = Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

=Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

RHS = Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

LHS ≠ RHS

Change of Order of Integration

In calculus, interchange of the order of integration is a methodology that transforms iterated integrals of functions into other.

Theorem : If S is a region of Type I, then

Integral Calculus Notes | EduRev

If S is a region of Type II, then

Integral Calculus Notes | EduRev

We should point out that, unlike the case of integrals over rectangles, there is only one order in which we can carry out the integrations. If S is a Type I region we have to integrate over y before we integrate over x and if S is a Type II region we have to carry out the integration over x before we integrate over y.

However, sometimes a region is both Type I and Type II. This does not mean that

Integral Calculus Notes | EduRev

because the right hand side doesn’t really make any sense. Rather, it means that we can prescribe the region S in two different ways

Integral Calculus Notes | EduRev

The preceding theorem applied to this situation simple says

Integral Calculus Notes | EduRev

In order to reverse the order of integration of an integral like

Integral Calculus Notes | EduRev

one therefore first has to Fig. out how to parameterize the region of integration 

Integral Calculus Notes | EduRev

as a Type II region; that is to say, one has to Fig. out what c, d, ψ1(y), and ψ2(y) are.

Example 6 : Find the value of Integral Calculus Notes | EduRev if the domain E integration is the triangle bounded by the straight lines y = x, y = 0 and x = 1.

How to take limit of y and x, we draw a strip parallel to y axis in the fig.

Now lower end of strip is on the line y = 0 and upper end of strip is on the line y = x So y varies from y = 0 to y = x.

For x, we pull the strip at the left end of shaded fig i.e. x = 0 and extend the strip i.e the right end of shaded region i.e. x = 1 So y varies from y = 0 to x.

To avoid integration of ey/x with respect to x, we use

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 7 : Evaluate the double integral Integral Calculus Notes | EduRev, when E is the domain which lies between two squares of sides 2 and 4, with center at the origin and sides parallel to the axes.

Domain E is not quadratic with respect to any axes but the straight lines x = -1, x = 1 divide it into four quadratic domains, E1, E2, E3, E4.

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

see the region E1, limit of x is from -2 to -1 and limit of y is from -2 to 2 Similarly find limits for other regions also.

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev
= Integral Calculus Notes | EduRev

= 2 cosh(4) - 2 cosh(2)

Example 8 : Evaluate Integral Calculus Notes | EduRev, over the rectangle R = [0, 1; 0, 1], where

Integral Calculus Notes | EduRev

For non-zero values of the function f(x, y), the domain R is divided into two domains A and B, we obtain

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 9 : Evaluate Integral Calculus Notes | EduRev, over the rectangle R = [0, 1 : 0, 2], where [x + y] denotes the greatest integer less than or equal to (x + y). We have, for (x, y) ∈ R.

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

The domain of integration R (i.e. region ABDE) is divided into three domains ABF, BCEF and CDE and [x + y] = 0 in the region ABF, [x + y] = 1 in region BCEF and [x + y] = 2 in the region CDE.

Therefore, we have

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= 3 [1 - 0] + 2 [2 - 0] 

= 3 + 4 = 7

Jacobian of Transformation

f(x, y)dx dy = f(r cosθ, r sinθ) Jdr dθ 

dx dy = Jdr dθ

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

which is known as Jacobian.

Jacobian of transformation form of (x, y) into (r, θ)

J = Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Change of Variables in Double Integrals

Let the functions x = X(u, v); y = Y(u, v) map in one-to-one manner a domain D in Cartesian coordinates (x, y) onto a domain D’ in the new coordinates u, v.

Let f(x, y) = f(X(u, v), Y(u, v)) = F(u, v)

Then Integral Calculus Notes | EduRev

where the Jacobian J = Integral Calculus Notes | EduRev

In case of a transformation from a Cartesian coordinate system (x, y) to polar coordinate system (r, θ); I f I = r, and hence dx dy = r dθ dr.

Change of Variables in Triple Integrals 

Let the functions

x = X(u,v,w), y = Y(u,v,w), z = Z(u,v,w)

map in one-to-one manner, a domain D in cartesian coordinates (x,y,z) onto a domain D’ in the new variables (u,v,w)

Let f(x,y,z) = f(X(u,v,w), Y(u,v,w), Z(u,v,w)) = F(u,v,w)

Then Integral Calculus Notes | EduRev

where the Jacobian J = Integral Calculus Notes | EduRev (matrix is same as given by eq. (1) of order 3)

Integral Calculus Notes | EduRev

The following two transformations because of their frequent occurrence, deserve special mention 

(i) Cylindrical polar coordinates

x = r cosθ, y = r sinθ, z = z Here the Jacobian J = r.

ii) Spherical polar coordinates. x = r sinθ cos ∅, y = r sinθ sin ∅, z = r cosθ 

Here the Jacobian J = r2 sinθ

Examples 10 : To integrate (x2 + y2) over the circle x2 + y2 = a2, we change to polar, x = r cosθ, y = r sinθ, so that Integral Calculus Notes | EduRev.

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev (integrating wrt r)

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

1. To evaluateIntegral Calculus Notes | EduRev over the positive quadrant of the ellipse x2/a2 + y2/b2 = 1, by putting x = au ⇒ d x = a du, and y = bv = > d y = bdv, we have to evaluate ab Integral Calculus Notes | EduRev dv, over the positive quadrant of the circle u2 + v2 = 1.

Integral Calculus Notes | EduRev

Again, changing to polars, by putting u = r cos 0, v = r sin 0, the integral becomes

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

put 1 + r2 = p2 ⇒ 2rdr = 2pdp (lim it: r = 0 ⇒ p = 1, r = 1⇒ p = √2 )

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

put p = √2 sint

dp = √2 cost dt

= Integral Calculus Notes | EduRev

Put p = √2 sin t 

dp = - √2 cost dt

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Note : The transformation could be effected in one step by putting x / a = rcosθ, y /b = rsinθ, then I J I = abr.

2. To evaluate ∫∫{ 2a2 - 2a( x + y) - ( x2 + y2) } dx dy over the circle x2 + y. + 2a(x + y) = 2a2, transform the origin to ( - a, - a ), by putting x + a = u, y + a = v, so t hat the integral becomes ∫∫( 4a2 - u. - v2) dudv, over the circle u2 + v2 = 4a2. Changing to polars , we get 8πa4.

Integral Calculus Notes | EduRev over the circle

x2 + y2 + 2a(x + y) = 2a

x+ 2ax + a.+ y2 + 2ay + a2 = 4a2 

(x + a)2 + (y + a)2 = 4a2 

put x + a = u, y + a = v 

So the integral becomes

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Changing into polar coordinates

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= 8πa4

Example 11 : Evaluate ∫ ∫y - x ) d x dy, over the region E in the xy-plane bounded by the straight lines y = x - 3 , y = x+1,3y + x = 5,3y + x = 7

It is difficult to evaluate the double integral directly; however a simple change of coordinates reduces the domain of integration into a rectangle with sides parallel to the axis.

Set y - x = u, 3y + x = v

so that

Integral Calculus Notes | EduRev

and Integral Calculus Notes | EduRev

The new domain is the rectangle R = [-3, 1; 5, 7] in uv-plane.

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 12 : Evaluate the integral Integral Calculus Notes | EduRev by passing on to the polar coordinates.

The integral in question is the double integral Integral Calculus Notes | EduRev dx dy over the region enclosed by the triangle y = 0, y = x, x = 1.

In polar coordinates, the equations of these lines are 0 = 0, 0 = rc/4. r cos 0 = 1, so that the domain of integration is 0 < 0 <π/4, 0 < r < sec 0.

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Dirichlet's Theorem 

The theorem states that

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

where the integral is extended to all +ve values of the variables x1, x2, ..., xn subject to the condition x1 + x2 + ..., x≤ 1.

Liouville's Extension of Dirichlet's Theorem

If x, y, z all positive such that 

h,≤ x + y + z ≤ h2

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Example 13 : Evaluate Integral Calculus Notes | EduRev where E is the region bounded by x = 0, y = 0, x + y = 1.

Now Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

For this type of integrals, two sets of substitutions are possible, which give an integral with constant limits. We discuss both these here.

First method. Put x + y = u, x = uv 

so that y = u(1 - v) and

Integral Calculus Notes | EduRev

The Jacobian vanishes when u = 0, i.e., when x = 0 = y. The origin of the xy-plane corresponds to the whole line u = 0 of the uv-plane, so that the correspondence ceases to be one-to-one. To exclude the origin of the xy-plane, we cut out the region along a line x = h parallel to the y-axis and consider the integral on the remaining domain E,, bounded by the lines 

v = 1, u = 1, uv = h

Integral Calculus Notes | EduRev

However, in the limit when h -> 0, this new region degenerates into the square bounded by v = 1, u = 1, v = 0, u = h 

Thus Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Second method. Put x = u, y = (1 - u)v, so that 

1 - x - y = 1 - u - ( 1 - u ) v = (1 - u)(1 - v )

and Integral Calculus Notes | EduRev

which vanishes for u = 1, so that the point (1, 0) in the xy-plane corresponds to the whole line u = 1 in the uv-plane. However, proceeding as in the first method, the new region becomes the square.

u = 0, v = 0, u = 1, v = 1

Thus Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 14 : Evaluate Integral Calculus Notes | EduRev dxdy, where E is the region bounded by the co-ordinate  axes and x + y = 1 in the first quadrant.

Taking x - y = u, x + y = v, so that x = 1/2(u+v), y = 1/2(v-u) and the Jacobian is 1/2.

Integral Calculus Notes | EduRev -------(1)

Integral Calculus Notes | EduRev

Now Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev   ----(2)

Hence, from (1) and (2), the required integral is zero.

Triple Integrals

Let f(x, y, z) be a function of three independent variables x, y and z defined at every point of a three-dimensional region V. Divide the region V into n elementary volumes δV1f δV2, δVn and let (xr, yr, zr) be any point inside the rth sub-division δVr. Find the sum

Integral Calculus Notes | EduRev

Then Integral Calculus Notes | EduRev

To extend definition of repeated integrals for triple integrals, consider a function f(x, y, z) and keep x and y constant and integrate with respect to z between limits in general depending upon x and y. This would reduce f(x, y, z) to a function of x and y only. Thus, let

Integral Calculus Notes | EduRev

Then in ∅ (x, y) we can keep x constant and integrate with respect to y between limits in general depending upon x this leads to a function of x alone, say

Integral Calculus Notes | EduRev

Finally ψ(x) is integrated with respect to x assuming that the limits for x are from a to b.

Thus

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

If we put f(x, y, z) = 1, then the volume

Integral Calculus Notes | EduRev

Use of definite triple integral in finding volume of 3D shape : 

If we have a 3D-shape in which two very-very closed point (P, Q) are taken as diagonal of a cuboid.Integral Calculus Notes | EduRev

Then volume of this elementary cuboid inside that 3D shape 

dV = dx dy dz

Integral Calculus Notes | EduRev taken over region

Integral Calculus Notes | EduRev

Example 15 : Evaluate the integral Integral Calculus Notes | EduRev

Let

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 16 : Evaluate Integral Calculus Notes | EduRev here V is the volume of the cube bounded by the coordinate planes and the planes x = y = z = a.

Here a column parallel to z-axis is bounded by the planes z = 0 and z = a.

Here the region S above which the volume V stands is the region ist he xy-plane bounded by the lines x = 0, x = a, y = 0, y = a.

Hence the given integral

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 17 : Integral Calculus Notes | EduRev where V is the closed region bounded by the cylinder z = 4 - x2 and the planes x = 0, y = 0, y = 2 and z = 0.

Here a column parallel to z-axis is bounded by the plane z = 0 and the surface z = 4 - x2 of the cylinder.

This cylinder z = 4 - x2 meets the z-axis, x = 0, y = 0, at (0, 0, 4) and the x-axis, y = 0, z = 0 at (2, 0, 0) in the given region.

Therefore, it is evident that the limits of integration for z are from 0 to 4 - x2, for y from 0 to 2 and for x from 0 to 2.

Here the given integral

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 18 : Evaluate Integral Calculus Notes | EduRevtaken over the region common to the surfaces x2 + y2 + z2 = a2, and x2 + y2 = ax

Integral Calculus Notes | EduRev

The region is bounded, above and below by the surfaces

Integral Calculus Notes | EduRev

and its projection on the xy-plane is the circular domain D = x2 + y2 ≤ ax.

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Changing to polars, the region D becomes the circle, r = a cosθ

0 ≤ θ ≤ π, 0 ≤ r ≤ a cosθ

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 19 : Evaluate Integral Calculus Notes | EduRev taken over the domain bounded by the cylinder x2 + y2 = 2ax, and the cone z2 = k2(x2 + y2).

The domain E is bounded above and below by the surface

Integral Calculus Notes | EduRev  and Integral Calculus Notes | EduRev

and its projection on the xy-plane is the circular domain D, x2 + y2 ≤ 2ax.

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Changing to polars,

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Change of Variables in A Triple Integral

Let f be continuous in a domain E bounded by a surface S in xyz-space.

Also let

x = x (X, Y, Z), y = y (X, Y, Z); z = z (X, Y, Z).

define three functions of three variables X, Y, Z, defined in a domain E, bounded by a surface S, in XYZ-space.

We suppose that these three functions with values 

x (X, Y, Z), y(X, Y, Z), z(X, Y, Z) 

(i) possess continuous first order partial derivatives at each point of E, and Sr 

(ii) transform E, into E and S1 into S. 

(iii) the transformation is one-one. 

(iv) the Jacobian

Integral Calculus Notes | EduRev

does not change sign at any point of E1, even though it may vanish at some points of S1 It will then be proved that

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

where x, y, z have to be replaced by their values in terms of X, Y, Z in f(x, y, z) in the integral on the right.

Example 20: Show that Integral Calculus Notes | EduRev taken over the tetrahedron bounded by the planes, x = 0, y = 0, z = 0, Integral Calculus Notes | EduRev

The given integral is same as

Integral Calculus Notes | EduRev

First Method . Let us put x + y + z = u, x + y = uv, x = uvw, i.e., std. substitution 

x = uvw, y = uv(1 - w), z = u(1 - v)

It may be seen that when x, y, z are positive and x + y + z < 1, then each of u, v, w lie between 0 and 1 and conversely. So the given region is fully described when 0 ≤ 1 , 0 ≤ v ≤ 1 , 0 ≤ w ≤ 1 .

Also, them

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Second Method . Put x = u, y = (1 - u)v, z = (1 - u) (1 - v)w so that 

1 - x - y - z = ( 1 - u)(1 - v)(1 - w)

and Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Example 21 : Evaluate Integral Calculus Notes | EduRev taken throughout the domain Integral Calculus Notes | EduRev

Change the variables x, y, z to X, Y, Z where x = aX, y = bY, z = cZ, 

so that ∂(x, y, z)/∂(X, Y, Z) = abc.

Thus Integral Calculus Notes | EduRev

taken throughout X2 + Y2 + Z2 ≤ 1.

Changing X, Y, Z to polar co-ordinates r, θ, φ so that X = r sinθ cosφ, Y = r sinθ sinφ, Z = r cosθ,

We have, since

∂(X, Y, Z)/∂(r, θ, φ) = r2 sin θ,

Integral Calculus Notes | EduRev

It is easily seen that to describe the whole region.

X2 + Y2 + Z2 ≤ 1.

r varies from 0 to 1; 0 varies from 0 to π ; φ varies from 0 to 2π.

Thus (r, 0, φ) varies in the rectangular parallelepiped.

[0, 1; 0, π; 0, 2π]

Integral Calculus Notes | EduRev

SURFACE AREA

Definition : The area of the surface 

x = f(u, v), y = g(u, v), z = h(u, v); (u, v) ∈ D is the double integral

Integral Calculus Notes | EduRev assuming that the functions f, g, h possess continuous first order partial derivatives in D and at no point of D.

Integral Calculus Notes | EduRev

Note : Here, we do not propose to derive the expression as given above by basing the derivation on some elementary notion of surface area. Let the definition as given appear very arbitrary, we outline some considerations to call forth reader’s faith in the same.

I. Let the surface be plane. We take 

x = u, y = v, z = 0 

where (u, v) ranges over a domain D in the XY-plane. In fact, the surface coincides with D in the present case.

We have

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Thus the definition of surface area as now given agrees with that of areas of plane regions.

II. Consider the surface given by z = h (x, y).

We take it as given by 

z = u, y = v, z = h(u, v) 

where (u, v) = (x, y) ∈ D ⊂ R3

We have

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

∴ surface area = Integral Calculus Notes | EduRev -----(1)

The reader acquainted with the elements of Differential Geometry knows that 

Integral Calculus Notes | EduRev are the direction ratios of the normal to the surface at any point (x, y, z) , so that if we suppose δσ to be an element of surface lying on the tangent plane at the point, the projection area δA of δσ on the XY-plane is given by

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

This suggests, on summation, the truth of (1). 

Note. The reader may easily show that

Integral Calculus Notes | EduRev

where E = Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Smooth surface.

Definition: A surface given by x = f(u, v), y = g(u, v), z = h(u, v); (u, v) ∈ D is said to be smooth, if f, g, h possess continuous first order partial derivatives at each point of D and at no point there of

Integral Calculus Notes | EduRev vanish simultaneously.

Example 22: The area of the surface of the paraboloid 2z = x2 + y2 which lies between the planes z = 0, z = 2 is ______

Here

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

The projection on the plane z = 2 is x2 + y2 = 4 or r = 2 which is a circle between θ = 0 and θ = 2π (changing to polars by putting x = r cos θ, y = r sinθ). Hence

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

=Integral Calculus Notes | EduRev

Example 23 : Find the area of the surface az = xy that lies inside the cylinder (x2 + y2)2 = 2a2xy.

We have az = xy.

Integral Calculus Notes | EduRev 

Surface area = Integral Calculus Notes | EduRev

=Integral Calculus Notes | EduRev changing into polars .(x+ y2 =r2 , x = r cosθ , y = r sinθ ) 

Also (x2 + y2)2 = 2a2xy becomes r2 = a2 sin 2θ curve is a symmetric about origin and see the loop of the curve lies between θ= 0 to π/2.

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev -------(1)

where 

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev ------(2)

By equation (i),

Integral Calculus Notes | EduRev

Example 24 : Compute the surface area S of the sphere x2 + y2 + z2 = a2

The surface area of the sphere is twice the surface are a of the upper half-sphere 

z = Integral Calculus Notes | EduRev

Now Integral Calculus Notes | EduRev

so that Integral Calculus Notes | EduRev

The domain of integration is the circle x2 + y2 = a2 on the xy-plane. 

Thus by formula (1), we have

Integral Calculus Notes | EduRev

On passing to polars, we have

Integral Calculus Notes | EduRev

Example 25 : Find the area of that part of the surface of the cylinder x2 + y2 = a2 which is cut out by the cylinder x2 + z2 = a2.

Integral Calculus Notes | EduRev

The Fig. shows 1/8th of the desired surface

The equation of the surface has the form

Integral Calculus Notes | EduRev 

so that Integral Calculus Notes | EduRev

Integral Calculus Notes | EduRev

The domain of integration is a quarter circle x2 + z2 = a2, x ≥ 0, z≥ 0, on the xz-plane. Thus by formula (3), we have

Integral Calculus Notes | EduRev

∴ S = 8a2.

Evaluation of Volumes

By Triple Integration

The triple integral

Integral Calculus Notes | EduRev

carried throughout a region E in space of three dimensions gives the volume of E.

By Double Integration Let C be the boundary of a region E of the xy-plane and let a cylinder be constructed by lines through the points of C parallel to z-axis. Then the volume of the cylinder enclosed between the surfaces

Integral Calculus Notes | EduRev 

is Integral Calculus Notes | EduRev  and can be easily seen. given by the double integral

Integral Calculus Notes | EduRev

Example 26: The volume enclosed by the surfaces x+ y2 = cz, x2 + y= 2ax, z = 0.

The limits of z are 0 to Integral Calculus Notes | EduRev The limits of y are from Integral Calculus Notes | EduRev and the limits of x are from 0 to 2a.

The value

Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

=Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Put a - x = a sin θ

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

=Integral Calculus Notes | EduRev

Example 27 : The sphere x2 + y2 + z2 = a2 is pierced by the cylinder (x2 + y2)2 = a2(x2 - y2); prove that the volume of the sphere that lies inside the cylinder is Integral Calculus Notes | EduRev

Here limits of z are from Integral Calculus Notes | EduRev and therefore 

the volume = Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Now the equation of the cylinder is 

(x2 + y2)2 = a2(x2 - y2)

Putting x = r cosθ, y = r sinθ, we get 

r2 = a2 cos2θ.

∴ The limits of r are from Integral Calculus Notes | EduRev and limits of 0 are from -π/4 to π/4. 

∴ The required volume

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

= Integral Calculus Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Extra Questions

,

Previous Year Questions with Solutions

,

Summary

,

Free

,

ppt

,

pdf

,

MCQs

,

Exam

,

Objective type Questions

,

past year papers

,

study material

,

Viva Questions

,

Important questions

,

shortcuts and tricks

,

practice quizzes

,

Integral Calculus Notes | EduRev

,

Integral Calculus Notes | EduRev

,

Sample Paper

,

Integral Calculus Notes | EduRev

,

mock tests for examination

,

video lectures

,

Semester Notes

;