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This mock test of Integral Calculus -8 for Mathematics helps you for every Mathematics entrance exam.
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QUESTION: 1

The area of the cardioid r = a (1 - cos θ) is given by:

Solution:

The given curve is

r= a(l - cos θ)

The required area A is given by

A = 2 x area ABOA

QUESTION: 2

The volume of the solid generated by the revolution of r = 2a cos θ about the initial line is given by:

Solution:

The equation of the given curve is

r= 2a cos θ,

which is a circle with centre (a, 0) and radius a.

Hence The required volume

QUESTION: 3

The surface area of the solid generated by the revolution of the curve x = a cos^{3}t, y = a sin^{3} t about x-axis is given by:

Solution:

The equation of the given curve is

x = a cos^{3} t,y = a sin^{3} t

Hence, the required surface area

QUESTION: 4

The length of the arc of the curve r = a e^{θ cotα} between the point whose radii vectors are given by r_{1} and r_{2} is:

Solution:

QUESTION: 5

The area of the ellipse is given by:

Solution:

QUESTION: 6

The segment o f the circle x^{2} + y^{2} = a^{2} cut off by the chord x = b (0 < b < a) revolves about the x-axis and generates the solid known as a segment of a sphere. The volume of this solid is:

Solution:

The curve bounded by x^{2} + y^{2} = a^{2} and chord x = b revolved about x-axis, then

The required volume is given by

QUESTION: 7

The surface area of the solid generated by the revolution of the curve r^{2} = a^{2} cos 2θ about a tangent at the pole is given by:

Solution:

Let P(r, θ) be a point on the lemniscate

r^{2} = a^{2} cos 2θ

Let OT be a taneent at the pole

QUESTION: 8

The area bounded by the curve y = sin x, x-axis and the lines x = 0, x = π is revolved about y-axis. The surface of revolution is equal to:

Solution:

The equation of the curve is y = sin x.

Hence The required surface area

Put cos x = t then sin x dx - dt

QUESTION: 9

The surface of the solid formed by the revolution o f the cardioid r = a (1 + cos θ) about the initial line is given by:

Solution:

The equation of the cardioid is

r = a (1 + cos θ)

The curve is symmetrical about the initial line. The required surface area

QUESTION: 10

The length of the are of the curve

x sin θ + y cos θ = f' (θ),

x cos θ - y cos θ = f " (θ)

is given by:

Solution:

The equations of the given curve are

x sin θ + y cos θ = f' (θ),

x cos θ - y cos θ = f " (θ)

Solving these equations, we get

Differentiating these w.r. to θ, we get

Hence the required length is given by

QUESTION: 11

The volume of the solid generated by the revolution of the cardioid r = a(1 + cos θ) about the initial line is given by

Solution:

The curve is r = a (1 + cos θ)

Required volume

QUESTION: 12

The ratio of areas bounded by the curve a^{4}y^{2} = x^{5}(2a -x) and the circle of radius a is:

Solution:

The area of curve =

Put x = 2 a sin^{2} θ

Then dx = 4a sinθ cosθ dθ

Hence, required ratio =

QUESTION: 13

By taking (a, 0) as the fixed point, the intrinsic equation of the astroid x^{2/3} + y^{2/3} = a^{2/3} is given by:

Solution:

The equation of the given curve is

Then

Since, (a, 0) is taken as a fixed point, therefore s increases as θ increases.

Hence

..(i)

QUESTION: 14

The area enclosed by the curve r^{2 }= a^{2} sinθ is given by:

Solution:

The equation of the given curve is

r^{2} = a^{2}sinθ

Required area

= Area of OBACO

= 2 x Area OBAO

QUESTION: 15

The volume of the solid generated by the revolution of the lemniscate r^{2 }= a^{2} cos2θ about the line is given by:

Solution:

The required volume

Thus

QUESTION: 16

If f(x) is continuous in [3, 7], then is equal to:

Solution:

We have

QUESTION: 17

The area bounded by the curves y^{2} = 9x, x - y + 2 = 0 is given by:

Solution:

The equations of the given curves are

y^{2 }= 9x ...(I)

x - y + 2 = 0 ...(II)

The curves (i) and (ii) intersect at A(1, 3) and B(4, 6) Hence The required area

QUESTION: 18

The area bounded by the curve y^{2 }= x^{3} and the line y = 2x is given by:

Solution:

The given curves are

y^{2} = x^{3} ...(i)

y = 2x ...(ii)

Solving (i) and (ii), we get

x = 0, x = 4

Hence the required area OPAQO

= 16/5

QUESTION: 19

The area bounded by the curve x^{2 }= y and y^{2} = x is given by:

Solution:

Given equations of the curves are

y^{2} = x ...(i)

x^{2} = y ...(ii)

Solving these equations, we get

x = 0, x = 1

Hence required area OPAQO

QUESTION: 20

The area bounded by the parabola y^{2} = 4ax and its latus rectum is given by

Solution:

The equation of the parabola is

y^{2} = 4 ax

We have to find the area OL'MLO

Hence Area OL'MLO = 2 • Area OMLO

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