Question

Find the volume and surface area of the solid generated by the revolution of the asteroid x^2/3+y^2/3=a^2/3 about the x-axis

Answer 1

Step-by-step explanation:

This is the answer.

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Answer 2

Given,

x^2/3+y^2/3=a^2/3

To Find,

The volume and surface area of the solid generated by the revolution of the asteroid.

Solution:

The given asteroid is x2/3+y2/3=a2/3 . First we will substitute the value of x as acos3t and y as asin3t i.e.

x=acos3t

y=asin3t

The given asteroid is symmetrical about the x axis.

Now, we will calculate dx/dt by differentiating with respect to.

dx/dt = d(acos3t)/dt

On differentiating, we get

dx/dt = −3acos2tsint

Similarly, we will calculate dy/dt by differentiating with respect to.

dy/dt = d(asin3t)/dt

On differentiating, we get

dy/dt=3asin2tcost

Now, we will calculate ds/dt, where ds/dt=$\sqrt{(dx/dt)2+(dy/dt)2}$

We will put the value of dx/dt and dy/dt here

Thus, ds/dt = $\sqrt{(−3acos2tsint)2+(3asin2tcost)2}$

We will calculate the squares of the terms.

ds/dt = $\sqrt{9a2cos4tsin2t+9a2sin4tcos2t}$

On further simplification, we get

ds/dt = 3asintcost $\sqrt{cos2t+sin2t}$

We know from trigonometric identities,

cos2t+sin2t = 1

Thus, ds/dt becomes;

ds/dt = 3asintcost

We will find the limits of t using the limits of x and y.

As the value of x is varying from −a to a.

We will find the limit of t using x = acos3t

Therefore, value of t when x is a is

a = acos3t

t = 0

Therefore, value of t when x is −a is

−a = acos3t

t = π

Thus, t is varying from 0 to π.

The surface area of the solid generated = ∫0π2πx(ds/dt).dt

On simplifying the integration, we get

The surface area of the solid generated =2∫0π22πx(ds/dt).dt

On putting the value of ds/dt and x, we get

The surface area of the solid generated=2∫0π/22πa.cos3t.3asintcostdt

Taking constants out of integration, we get

The surface area of the solid generated = 12πa2∫0π/2sintcos4tdt

Let cost be v, so −asint will become dv. The limits become cos(0) = 1 to cosπ2=0.

Therefore,

The surface area of the solid generated =12πa2∫10−v4dv

On integration, we get

The surface area of the solid generated

= −12πa2(v5/5)01=−12πa2(0−1/5)=12/5πa2

Thus, the required surface area is  12/5πa2.

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