Integral Calculus -8 - Mathematics MCQ

The given curve is
r= a(l - cos θ)
The required area A is given by
A = 2 x area ABOA

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

The equation of the given curve is
x = a cos³ t,y = a sin³ t

Hence, the required surface area

The curve bounded by x² + y² = a² and chord x = b revolved about x-axis, then
The required volume is given by

Let P(r, θ) be a point on the lemniscate
r² = a² cos 2θ
Let OT be a taneent at the pole

The equation of the curve is y = sin x.
Hence The required surface area



Put cos x = t then sin x dx - dt

The equation of the cardioid is
r = a (1 + cos θ)
The curve is symmetrical about the initial line. The required surface area

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

The curve is r = a (1 + cos θ)
Required volume




 

The area of curve = 
Put x = 2 a sin² θ
Then dx = 4a sinθ cosθ dθ




Hence, required ratio = 

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)

The equation of the given curve is 
r² = a²sinθ
Required area 
= Area of OBACO
= 2 x Area OBAO 

The required volume

Thus

We have 

The equations of the given curves are
y² = 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

The given curves are
y² = x³ ...(i)
y = 2x ...(ii)
Solving (i) and (ii), we get 
x = 0, x = 4
Hence the required area OPAQO



= 16/5

Given equations of the curves are
y² = x ...(i)
x² = y ...(ii)
Solving these equations, we get 
x = 0, x = 1
Hence required area OPAQO

The equation of the parabola is 
y² = 4 ax
We have to find the area OL'MLO
Hence Area OL'MLO = 2 • Area OMLO