   Chapter 15.7, Problem 20E

Chapter
Section
Textbook Problem

Use cylindrical coordinates.20. Evaluate ∭ E ( x   −   y )   d V , where E is the solid that lies between the cylinders x2 + y2 = 1 and .x2 + y2 = 16, above the xy-plane, and below the plane z = y + 4.

To determine

To evaluate: The given triple integral by using cylindrical coordinates.

Explanation

Given:

The function is f(x,y,z)=xy .

The region E lies between the cylinders x2+y2=1,x2+y2=16 above the xy plane and below the plane z=y+4 .

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) corresponds to the rectangular coordinates (x,y,z) is,

r=x2+y2θ=tan1(yx)z=z

Calculation:

From the given conditions, it is observed that r varies from 1 to 4, θ varies from 0 to 2π and z varies from 0 to rsinθ+4 . Use the formula mentioned above to change the given problem into cylindrical coordinates. Then, by the equation (1), the value of the given triple integral is,

Ef(x,y,z)dV=1402π0rsinθ+4(xy)(r)dzdθdr=1402π0rsinθ+4(rcosθrsinθ)(r)dzdθdr=1402π0rsinθ+4(r2cosθr2sinθ)dzdθdr

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

1402π0rsinθ+4(r2cosθr2sinθ)dzdθdr=1402π(r2cosθr2sinθ)[z]0rsinθ+4dθdr=1402π(r2cosθr2sinθ)(rsinθ+40)dθdr=1402π(r3cosθsinθ+4r2cosθr3sin2θ4r2sinθ)dθdr=1402π(r3cosθsinθ+4r2cosθr3sin2θ4r2sinθ)dθdr

On further simplification, the value of triple integral becomes,

1402π0rsinθ+4(r2cosθr2sinθ)dzdθdr=1402π(r32sin2θ+4r2cosθr3sin2θ4r2sinθ)dθdr=1402π(r32sin2θ+4r2cosθr32(1cos2θ)4r2sinθ)dθdr=1402π(r32sin2θ+4r2cosθr32+r32cos2θ4r2sinθ)dθdr

Integrate with respect to θ and apply the limit of it

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