Integral Calculus NAT Level - 2 - Physics MCQ

This is a horizontal ellipse 

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

Total area in first quadrant 
Total area 
Hence, value of λ = 4
The correct answer is: 4

Let us first calculate the area included between two cardioids

Shaded area = 4 × Area OL in first quadrant



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

Hence, required area

The correct answer is: 2.428

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

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.

The correct answer is: 0.333

Slope of this curve is equation of the curve 

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


The correct answer is: 0.375

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.
The correct answer is: 4

Hence, the required area  

Taking Mod sign we get 

The correct answer is: 0.5

The correct answer is: 2.666

Let 
Then   keeping x constant when

Hence, θ varies from 0 to x.

Changing the order of integration with the help of figure

The correct answer is: 0.894

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

Now unshaded Area = Whole Cardioid - Shaded Area

The correct answer is: 1.777