1. ## Surface integral

Evaluate the surface integral:

$\iint_S (1+x^2+y^2)dS$ where S is part of the surface of the paraboloid $z=1-x^2-y^2$ in the second octant.

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First of all, I hope the second octant is where x is negative but y and z are positive? (well, or 0).

Second,

Let $f(x,y,z)=z+x^2+y^2-1$ then $f=0$ and $\nabla f = <2x,2y,1>$ which is normal to the surface S.
EDIT: so $|\hat{n}\cdot\hat{k}|=\frac{1}{\sqrt{4x^2+4y^2+1}}$

Then in polar coords, in the 2nd octant, we have:
$0\le r \le 1$ since the shadow D on the xy-plane is a unit circle and,
$\pi/2 \le \theta \le \pi$ in the 2nd oct.

So the integral is:
$\int_{\pi/2}^\pi \int_0^1 (1+r^2)\sqrt{4r^2+1} r dr d\theta$. Right so far?

Now the problem is, I don't see any simple way to integrate this... I don't think u-substitution will help me here, but could be wrong.

2. Originally Posted by scorpion007
Evaluate the surface integral:

$\iint_S (1+x^2+y^2)dS$ where S is part of the surface of the paraboloid $z=1-x^2-y^2$ in the second octant.

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First of all, I hope the second octant is where x is negative but y and z are positive? (well, or 0).

Second,

Let $f(x,y,z)=z+x^2+y^2-1$ then $f=0$ and $\nabla f = <2x,2y,1>$ which is normal to the surface S.
EDIT: so $|\hat{n}\cdot\hat{k}|=\frac{1}{\sqrt{4x^2+4y^2+1}}$

Then in polar coords, in the 2nd octant, we have:
$0\le r \le 1$ since the shadow D on the xy-plane is a unit circle and,
$\pi/2 \le \theta \le \pi$ in the 2nd oct.

So the integral is:
$\int_{\pi/2}^\pi \int_0^1 (1+r^2)\sqrt{4r^2+1} r dr d\theta$. Right so far?

Now the problem is, I don't see any simple way to integrate this... I don't think u-substitution will help me here, but could be wrong.
Because you have "r dr", the first thing I would do is let $u= 4r^2+ 1$ to simplify that square root. du= 8r dr or (1/8)d= rdr. $\sqrt{4r^2+ 1}= u^{1/2}$. Since $4r^2+ 1= u$, $4r^2= u- 1$ and $r^2= u/4- 1/4$, $r^2+ 1= u/4+ 3/4$.
$\int_0^1 (1+r^2)\sqrt{4r^2+ 1} rdr= \int_1^5(\frac{u}{4}+ \frac{3}{4})u^{1/2}du= \int_1^5(\frac{1}{4}u^{3/2}+ \frac{3}{4}u^{1/2})du$

3. ah, thank you. I think I can solve it now