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# Integral Calculus NAT Level - 2

## 10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Integral Calculus NAT Level - 2

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This mock test of Integral Calculus NAT Level - 2 for Physics helps you for every Physics entrance exam. This contains 10 Multiple Choice Questions for Physics Integral Calculus NAT Level - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Integral Calculus NAT Level - 2 quiz give you a good mix of easy questions and tough questions. Physics students definitely take this Integral Calculus NAT Level - 2 exercise for a better result in the exam. You can find other Integral Calculus NAT Level - 2 extra questions, long questions & short questions for Physics on EduRev as well by searching above.
*Answer can only contain numeric values
QUESTION: 1

### Show that the area common to the ellipses a2x2 + b2y2 = 1, b2x2 + a2y2 = 1. when 0 < a < b is   Find the value of λ.

Solution:

This is a horizontal ellipse

This is a vertical ellipse
Total area = 4 x Area of OABC

Total area
Hence, value of λ = 4

*Answer can only contain numeric values
QUESTION: 2

### If the area shaded in the given figure is of the form λa2. Find the value of λ given the cardioids are r = a(1 + cosθ) and r = a(1 – cosθ) and a given circle r = a.

Solution:

Let us first calculate the area included between two cardioids

Now, this shaded portion lies entirely inside the circle r = a with area πa2

Hence, required area

*Answer can only contain numeric values
QUESTION: 3

### If ∭R xyz dx dy dz is solved using cylindrical coordinate where R is the region bounded by the planes x = 0, y = 0, z = 0, z = 1 & x2 + y2 = 1 then what is the value of that integral?

Solution:

x2 + y2 = 1 → ρ varies from 0 to 1 substituting x = ρ cos ∅, y = ρ sin ∅, z = z

z varies from 0 to1, x = 0, y = 0 → ∅ varies from 0 to π / 2

thus the given integral is changed to cylindrical polar given by

put sin ∅ = t, dt = cos ∅

t varies from 0 to 1

*Answer can only contain numeric values
QUESTION: 4

Find the area of one loop of the curve  If the area is of the form λπ.  Find the value of λ.

Solution:

The equation of the curve is

Turn the initial line through an angle π/18 and putting θ as   the above equation reduces to r = 2 cos 3θ and the tracing of this curve is as in figure.

Also for r = 2 cos 3θ when r = 0 we get cos 3θ = 0 or  and these is also symmetry about the initial line.

*Answer can only contain numeric values
QUESTION: 5

If the whole area of the curve given by the equation x = a cos3 t,  y = b sin3 t or   is of the form λπab.  Find the value of λ.

Solution:

Slope of this curve is equation of the curve

Total Area = 4 × area OABO
This is of parametric form

*Answer can only contain numeric values
QUESTION: 6

If the area between the curve x(x2 + y2) = a(x2 – y2) and its asymptote is A1 and the area of the loop is A2 Then value of A1 + A2 = λ·a2. Find the value of λ.

Solution:

The curve is symmetrical about x-axis. The loop is situated between lines x = 0 and x = a. The line x = a is asymptote of the curve,

We have,

For any point on arc OLA

For any point on arc OMB

Area between curve and its asymptotes

Area of the loop is given as

Hence, λ = 4.

*Answer can only contain numeric values
QUESTION: 7

Find the area lying outside the circle r = 2acosθ and inside the cardioid r = a(1 + cosθ). This is of form λπa2. Find value of λ.

Solution:

Hence, the required area

Taking Mod sign we get

*Answer can only contain numeric values
QUESTION: 8

Evaluate  where R is the region in the first quadrant that is outside the circle r = 2 and inside the cardioid r = 2(1 + cosθ).

Solution:

*Answer can only contain numeric values
QUESTION: 9

The value of  sin x sin-1 (sin x sin y) dxdy is

Solution:

Let
Then   keeping x constant when

Hence, θ varies from 0 to x.

Changing the order of integration with the help of figure

*Answer can only contain numeric values
QUESTION: 10

If the ratio of the two parts into which the parabola 2a = r(1 + cosθ) divides the area of the cardiod r = 2a(1 + cosθ)  is of the form  Find the value of β/α.

Solution:

Solving the given equation

Therefore shaded area = area OLMN
= 2 × area OMNO
2(area OMN + area ONO)

By putting  in the second integral and by putting θ/2 =  u in the first integral