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### 1 Answer

By evaluation theorem

`int int_Sf(x,y,z)ds=int int_{x^2+y^2=4} (x^2+y^2+(y+4)^2)sqrt(1+((delz)/(delx))^2+((delz)/(dely))^2)dA`

`since`

`z=y+4 =>(delz)/(delx)=0,(delz)/(dely)=1`

`therefore`

`=int int_{x^2+y^2=4)(x^2+y^2+(y+4)^2)sqrt(1+1)dA`

`` Since we have to integrate inside `x^2+y^2=4` . Let us convert it into polar coordinate

`x=rcos(theta)`

`y=rsin(theta)`

`=> r^2(cos^2(theta)+sin^2(theta))=4`

`=> r^2=4`

`=>r=2```

`So`

`int int_{x^2+y^2=4}(x^2+y^2+(y+4)^2)sqrt(2)dA`

`= sqrt(2)int_{theta=0}^{theta=2pi}int_{r=0}^{r=2}(r^2+(rsin(theta)+4)^2)r dr d theta`

`=sqrt(2)int_0^(2pi) int_0^2(r^2(1+sin^2(theta))+8rsin(theta)+16)rdr d theta`

`=sqrt(2)int_0^(2pi)((1+sin^2(theta))2^4/4+8sin(theta)2^2/2+16xx 2^2/2)d theta`

`=4sqrt(2)int_0^(2pi)(1+sin^2(theta)+4sin(theta)+8)d theta`

`=4sqrt(2)int_0^(2pi)(9+1/2-(1/2)cos(2theta)+4sin(theta))d theta`

`=4sqrt(2){(19theta)/2-(1/4)sin(2theta)-4cos(theta)}_0^(2pi)`

`=76sqrt(2)pi`

(So r changes from 0 t0 2 and `theta` changes from 0 to `2pi`.)

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