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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Evaluate the surface integral.

20. ∫∫s (x2 + y2 + z2) dS, S is the part of the cylinder x2 + y2 = 9 between the planes z = 0 and z = 2, together with its top and bottom disks

To determine

To evaluate: The value of S(x2+y2+z2)dS.

Explanation

Given:

The equation of the cylinder is y2+z2=9 and planes are z=0 and z=2.

Formula used:

Sf(x,y,z)dS=Df(r(u,v))|ru×rv|dA (1)

rθ=xθi+yθj+zθk (2)

rz=xzi+yzj+zzk (3)

S consists of three surfaces such are S1 be the lateral surface, S2 be the top disk and S3 be the bottom disk.

On S1 surface let x=3cosθ, y=3sinθ, z=z and limits 0z2 and 0θ2π.

r(θ,z)=3cosθi+3sinθj+zk

To find rθ, differentiate the above equation with respect to θ.

Substitute 3cosθ for x, 3sinθ for y and z for z in equation (2),

rθ=(3cosθ)θi+(3sinθ)θj+(z)θk=3(sinθ)i+3(cosθ)j+0k=3sinθi+3cosθj

To find rz, differentiate the above equation with respect to z.

Substitute 3cosθ for x, 3sinθ for y and z for z in equation (3),

rz=(3cosθ)zi+(3sinθ)zj+(z)zk=0i+0j+1k=1k

Find rθ×rz as follows,

rθ×rz=(3sinθi+3cosθj)×(1k)=|ijk3sinθ3cosθ0001|=(3cosθ0)i(3sinθ0)j+(00)k=3cosθi+3sinθj

Find |rθ×rz|.

|rθ×rz|=|3cosθi+3sinθj|=(3cosθ)2+(3sinθ)2=9cos2θ+9sin2θ=9(cos2θ+sin2θ)

Simplify the equation.

|rθ×rz|=9(1)=9{cos2θ+sin2θ=1}=3

Thus, |rθ×rz|=3.

Find the value of S1(x2+y2+z2)dS.

Rewrite equation (1) as follows.

S1(x2+y2+z2)dS=D(x2+y2+z2)(|rθ×rz|)dA

Apply limits and 9 for x2+y2 and 3 for |rθ×rz|,

S1(x2+y2+z2)dS=02π02(9+z2)(3)dzdθ=302πdθ02(9+z2)dz=3(2π)02(9+z2)dz=3(2π)[9z+z33]02

Simplify the equation.

S1(x2+y2+z2)dS=3(2π)[(18+83)0]=3(2π)(54+83)=3(2π)(623)=124π

On the surface S2, consider z=2 plane.

Let x=rcosθ, y=rsinθ, z=2 and limits 0r3 and 0θ2π.

r(θ,r)=rcosθi+rsinθj+2k (4)

To find rθ, differentiate the above equation with respect to θ.

Substitute rcosθ for x, rsinθ for y and 2 for z in equation (2),

rθ=(rcosθ)θi+(rsinθ)θj+(2)θk=r(sinθ)i+r(cosθ)j+0k=rsinθi+rcosθj

To find rr, differentiate the above equation with respect to r.

Modify equation (3) as,

rr=xri+yrj+zrk (5)

Substitute rcosθ for x, rsinθ for y and 2 for z,

rr=(rcosθ)ri+(rsinθ)rj+(2)rk=cosθi+sinθj+0k=cosθi+sinθj

Find rθ×rr.

rθ×rr=(rsinθi+rcosθj)×(cosθi+sinθj)=|ijkrsinθrcosθ0cosθsinθ0|=(00)i(00)j+(rsin2θrcos2θ)k=r(sin2θ+cosθ)k

rθ×rr=rk

Find |rθ×rr|.

|rθ×rr|=|rk|=(r)2=r2=r

To find S2(x2+y2+z2)dS modify equation (1) as follows

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