   Chapter 15.7, Problem 26E

Chapter
Section
Textbook Problem

Use cylindrical coordinates.26. (a) Find the volume of the solid that the cylinder r = a cos θ cuts out of the sphere of radius a centered at the origin.(b) Illustrate the solid of part (a) by graphing the sphere and the cylinder on the same screen.

(a)

To determine

To find: The volume of the given solid by using cylindrical coordinates.

Explanation

Given:

The region E is the cylinder r=acosθ cuts out of the sphere of radius a centered at origin.

Formula used:

If f is a cylindrical region E given by h1(θ)rh2(θ),αθβ, u1(rcosθ,rsinθ)zu1(rcosθ,rsinθ) where 0βα2π , then,

Ef(x,y,z)dV=αβh1(θ)h2(θ)u1(rcosθ,rsinθ)u2(rcosθ,rsinθ)f(rcosθ,rsinθ,z)rdzdrdθ (1)

The cylindrical coordinates (r,θ,z) corresponding to the rectangular coordinates (x,y,z) is,

r=x2+y2θ=tan1(yx)z=z

Calculation:

Equation of the sphere of radius a is x2+y2+z2=a2 .

Thus, the value of z is obtained below.

z2=a2x2y2z=±a2x2y2z=±a2r2

Therefore, r varies from 0 to acosθ , and θ varies from π2 to π2 and z varies from a2r2 to a2r2 .

Use the formula mentioned above to change the given problem into cylindrical coordinates. Then, by the equation (1), the volume of the given solid is computed as follows.

Ef(x,y,z)dV=π2π20acosθa2r2a2r2(r)dzdrdθ=π2π20acosθa2r2a2r2rdzdrdθ

Integrate with respect to z and apply the limit of it.

π2π20acosθa2r2a2r2rdzdrdθ=π2π20acosθr[z]a2r2a2r2drdθ=π2π20acosθr(a2r2(a2r2))drdθ=π2π20acosθr(a2r2+a2r2)drdθ=π2π20acosθ2ra2r2drdθ

Integrate with respect to r and apply the limit of it. For that substitute t=r2,dt=2rdr .

π2π20acosθa2r2a2r2rdzdrdθ=π2π2[2(a2t)323]0acosθdt=π2π2[2(a2r2)323]0acosθdθ=π2π2[2(a2(acosθ)2)3232(a2(0)2)323]dθ=13π2π2(2a3(1cos2θ)322a3)dθ

On further simplification, the required integral value is obtained

(b)

To determine

To illustrate: The given solid mentioned in part (a) by using graphing calculator.

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