Question

Surface integral:
\[int_{s}^{}int_{}^{}\frac{ds}{\sqrt{z-y+1}}\]
\[S: z=\frac{1}{2}x^{2}+y, 0\leq x\leq 1,0\leq y\leq 1 \]

Answer 1

so I used (x,y,g(x,y)) with g(x,y)=1/2x^2 +y

Answer 2

I stopped paying attention in class when they started surface integral

Answer 3

I guess I did too

Answer 4

I got:
\[\int\limits_{0}^{1}\int\limits_{0}^{1}\frac{\sqrt{4x^{2}+2}}{\sqrt{\frac{1}{2}x^{2}+1}}dxdy\]

Answer 5

Isn't it like line integral but instead of multiplying by arc lenght equation , we multiply by surface area equation

Answer 6

I think so, and you now have to variables to integrate of course.

Answer 7

You can get right of the y in mine of course:
\[\int\limits_{0}^{1}\frac{\sqrt{4x^{2}+2}}{\sqrt{\frac{1}{2}x^{2}+1}}dx\]

Answer 8

So it's down to a 1D integral now, If I am correct so far.

Answer 9

Should it be
\[\int\limits_{0}^{1}\frac{\sqrt{4x^{2}+1}}{\sqrt{\frac{1}{2}x^{2}+1}}dx\]?

Answer 10

No it's +2

Answer 11

oh, I see

Answer 12

forgot the (dz/dy)^2 + (dz/dy)^2+1
forgot about 1

Answer 13

right that's what I used.

Answer 14

I'm confused because wolfram alpha says this is not a nice integral: http://www.wolframalpha.com/input/?i=integrate+%28sqrt%284x^2%2B2%29%29%2F%28sqrt%28%281%2F2%29x^2%2B1%29%29+ {x%2C0%2C1}
But the answer is supposedly sqrt(2). So maybe I already made a mistake somewhere?

Answer 15

dz/dx = x --> x^2
\[\int _0^1\int _0^1\frac{\text{Sqrt}\left[x^2+2\right]}{\text{Sqrt}\left[\frac{1}{2}x^2+1\right]}dxdy\]

Answer 16

=\[\sqrt{2}\]

Answer 17

so the 4 is wrong?

Answer 18

Yes. I think you make error differentiating 1/2 x^2

Answer 19

oh, i see it

Answer 20

Thanks

Answer 21

you are welcome, S Int always gave me trouble as well