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Show $\int \int_{S} (x^2 +y^2) d\sigma = \frac{9\pi}{4}$ where $S = \left\{(x,y,z) : x>0, y>0, 3>z>0, z^2 =3(x^2+y^2)\right\}$.

We have the formula $\int \int_{S} f(x,y,z) = \int \int_{D} f(x, y, g(x,y)) \sqrt{(\frac{\partial z }{\partial x})^2 + (\frac{\partial z }{\partial y})^2+1}dA$. I figured I could rewrite the constraint on $z$ as $z = \sqrt{3(x^2+y^2)}$ which gives $ \frac{\partial z }{\partial x}=\frac{\sqrt{3}x}{\sqrt{(x^2+y^2)}}$ and $ \frac{\partial z }{\partial y}=\frac{\sqrt{3}y}{\sqrt{(x^2+y^2)}}$. Plugging this in to the formula gives $\int \int 2(x^2+y^2)$ after you simplify. I thought both integrals went from $0$ to $\sqrt{3}$, but this clearly doesn't give the correct answer. Where am I messing up?

user126004
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1 Answers1

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$\int \int_R 2(x^2+y^2)$ seems to be right. now you can solve this as:
$$\int_0^{\pi/2} \int_0^{\sqrt{3}} 2(r^3)drd\theta$$ this gives $9\pi/4$

ketan
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  • Ok, so I see that you converted to polar coordinates and I understand why $r$ goes from $0$ to $\sqrt{3}$, but why does $\theta$ go from $0$ to $\frac{\pi}{2}$? – user126004 Apr 08 '14 at 03:22
  • Wait, now I understand. It's because we are only integrating in the first quadrant, where $x,y>0$, i.e., $0 \leq \theta \leq \frac{\pi}{2}$. Thanks! – user126004 Apr 08 '14 at 03:24