   Chapter 15.7, Problem 19E

Chapter
Section
Textbook Problem

Use cylindrical coordinates.19. Evaluate ∭ E ( x   +   y   +   z )   d V , where E is the solid in the first octant that lies under the paraboloid z = 4 – x2– y2.

To determine

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

Explanation

Given:

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

The region E lies under the paraboloid z=4x2y2 in the first octant.

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:

Solve the given equation

z=4x2y24x2y2=04r2=0r=2

Therefore, r varies from 0 to 2. Since the given lies in the first octant, θ varies from 0 to π2 and z varies from 0 to 4r2 . 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=020π204r2(x+y+z)(r)dzdθdr=020π204r2(rcosθ+rsinθ+z)(r)dzdθdr=020π204r2(r2cosθ+r2sinθ+zr)dzdθdr

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

020π204r2(r2cosθ+r2sinθ+zr)dzdθdr=020π2[zr2cosθ+zr2sinθ+z2r2]04r2dθdr=020π2[((4r2)r2cosθ+(4r2)r2sinθ+(4r2)2r2)((0)r2cosθ+(0)r2sinθ+(0)2r2)]dθdr=020π2((4r2cosθr4cosθ)+(4r2sinθr4sinθ)+(16+r48r2)r2(0+0+0))dθdr=020π2[((4r2r4)cosθ+(4r2r4)sinθ+16r2+r528r32)]dθdr

Integrate with respect to θ and apply the limit of it

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