Question

Evaluate the surface integral.

double integral ydS

S is the part of the paraboloid
y = x2 + z2
that lies inside the cylinder
x2 + z2 = 1

Answer 1

This is a problem where the paraboloid projects a circle of radius 1 (represented by the region D) on the x-z plane.
y=0 when x=z=0
y=1 when x²+z²=1

So let
I=∫∫_DydS
It may be changed to cylindrical coordinates
where dS=rdrdθ, and using
y=x²+z²=r²,
I=∫[0,2π]∫[0,1]r²*rdrdθ
=2π[r^4/4][0,1] after separating r & θ
=2πr^4/4
=πr^4/2

Answer 2

what is r ? the answer doesn't have r in it

Answer 3

x^2 + z^2 = r^2 is a circle with radius r

read what mathmate wrote for you.