Asked by rebecca
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
Answers
Answered by
MathMate
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=∫∫<sub>D</sub>ydS
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
y=0 when x=z=0
y=1 when x²+z²=1
So let
I=∫∫<sub>D</sub>ydS
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
Answered by
rebecca
what is r ? the answer doesn't have r in it
Answered by
Damon
x^2 + z^2 = r^2 is a circle with radius r
read what mathmate wrote for you.
read what mathmate wrote for you.
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