Show that the area common to the ellipses a2x2 + b2y2 = 1, b2x2 + a2y2 = 1. when 0 < a < b is Find the value of λ.
This is a horizontal ellipse
This is a vertical ellipse
Total area = 4 x Area of OABC
Total area in first quadrant
Hence, value of λ = 4
The correct answer is: 4
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.
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
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?
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
Find the area of one loop of the curve If the area is of the form λπ. Find the value of λ.
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
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.
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
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.
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,
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
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.
Hence, the required area
Taking Mod sign we get
The correct answer is: 0.5
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).
The correct answer is: 2.666
The value of sin x sin-1 (sin x sin y) dxdy is
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
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.
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