   Chapter 10.3, Problem 89E

Chapter
Section
Textbook Problem

# Volume In Exercises 89 and 90, find the volume of the solid formed by revolving the region bounded by the graph of the parametric equations about the x-axis. Use the result of Exercise 77.) x = 6 cos θ , y = 6 sin θ

To determine

To calculate: The volume of the solid formed by revolving the region around x-axis where the region is bounded by the parametric equations, x=6cosθ,y=6sinθ.

Explanation

Given:

The parametric equations, x=6cosθ,y=6sinθ

Formula used:

If y is a continuous function of x on the interval axb, where x=f(t) and y=g(t), then,

|abydx|=|t1t2g(t)f(t)dt| where, f(t1)=a and f(t2)=b, and both g and f are continuous on [t1,t2].

And, the volume of the region when the curve is revolved around x-axis is given by, πaby2dx.

Calculation:

Consider the equation of the asteroid, x=6cosθ,y=6sinθ.

Assume that f(θ)=x and g(θ)=y.

Consider the formula,

πaby2dx=πabg2(θ)f(θ)dθ.

Now put y=y(θ) and dxdθ=f(θ) in the formula πaby2dx.

This gives, πaby2dx=πabg2(θ)f(θ)dθ.

Now use the formula, π|01y2dx|=π|abg2(θ)f(θ)dθ| to find the volume of the region.

So, differentiate f(θ) with respect to θ.

Then, f(θ)=6sinθ

Therefore, the volume of the region is given by, π|ab(6sinθ)2(6sinθ)dθ|.

Since, the figure revolves around x-axis.

Now for the upper limit and lower limits of the integral. The curves can varying θ from 0 to π because the revolution is around x-axis. If it has been around y-axis, θ would have varied from π2 to π2

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