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In exam it was asked to show that

$$\iint_S x^2+y^2 dA = 9 \pi /4$$ for $$S = {\{(x, y, z) | x>0, y>0,3>z>0, z^2 = 3(x^2 + y^2)}\}$$

I have tried many times but I don't get the $9 \pi /4$.

$$\begin{align} \iint_S\sqrt{1+f_x^2+f_y^2}\,dA &=\int_0^{\sqrt3}\int_0^{2\pi} r^2\sqrt{1+36r^2}\,r\,d\theta\,dr\\ &=2\pi\,\int_0^\sqrt3 r^3\sqrt{1+36r^2}\,dr = 2 \pi \dfrac{\left(36r^2+1\right)^\frac{3}{2}\left(54r^2-1\right)}{9720}\Bigg|_0^{\sqrt{3}} \\&= 2 \pi\dfrac{161{\cdot}109^\frac{3}{2}+1}{9720} \ne 9 \pi/4 \end{align}$$

Where I did it wrong?

Emma
  • 535

2 Answers2

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I am going try to reproduce your answer (at least how I think that you got there), so we can see where it went wrong.

Let $$ x=r\cos\theta, \quad y = r\sin\theta. $$ Then $$ z^2 = 3(x^2+y^2) = 3r^2 \quad \Rightarrow \quad z = r\sqrt{3} $$ where the final implication follows since both $z$ and $r$ are non-negative.

Hence $$ \Psi(r, \theta) = (r\cos \theta, r\sin\theta, r\sqrt{3}), $$ parameterizes our surface, but we must also find the correct intervals for $r$ and $\theta$. First, we have $$ 0 < z < 3 \quad \Leftrightarrow \quad 0 < r\sqrt{3} < 3 \quad \Leftrightarrow \quad 0 < r < \sqrt{3}. $$ Second, we know that $x,y > 0,$ so that they lie in the first quadrant, and we can conclude that $0 < \theta < \frac{\pi}{2}$.

Now it is only left to compute the surface element $dA$ before we can attack the integral, which can be done via $$ dA = \left|\frac{\partial \Psi}{\partial r} \times \frac{\partial \Psi}{\partial \theta}\right| \, dr\,d\theta. $$ The formula $\sqrt{1+f_x^2+f_y^2}$ that you seem to have used works only if you have a parameterization of the form $(x,y,f(x,y)),$ which is not the case in our case here. We have $$ \frac{\partial \Psi}{\partial r} = (\cos\theta, \sin\theta, \sqrt{3}), $$ and $$ \frac{\partial \Psi}{\partial \theta} = (-r\sin\theta, r\cos\theta, 0), $$ so \begin{align} \frac{\partial \Psi}{\partial r} \times \frac{\partial \Psi}{\partial \theta} &= \begin{vmatrix} e_1 & e_2 & e_3\\ \cos\theta & \sin\theta & \sqrt{3}\\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix}\\[0.2cm] &= e_1(0-\sqrt{3}r\cos\theta) + e_2(-\sqrt{3}r\sin\theta-0) + e_3(r\cos^2\theta+r\sin^2\theta)\\[0.1cm] &= (-\sqrt{3}r\cos\theta, -\sqrt{3}r\sin\theta, r), \end{align} and thus \begin{align} \left|\frac{\partial \Psi}{\partial r} \times \frac{\partial \Psi}{\partial \theta}\right| &= \sqrt{(-\sqrt{3}r\cos\theta)^2 + (-\sqrt{3}r\sin\theta)^2+r^2}\\ &=\sqrt{3r^2\cos^2\theta+3r^2\sin^2\theta+r^2} = r\sqrt{4} = 2r. \end{align} We have thus found your error - your surface element was wrong!

To complete the answer, we have \begin{align} \iint_S x^2+y^2 \, dA &= \int_0^{\pi/2}\int_0^{\sqrt{3}} r^2 \cdot 2r \, dr\, d\theta = \pi \int_0^{\sqrt{3}} r^3 \, dr = \frac{\pi}{4} [r^4]_0^{\sqrt{3}} = \frac{9\pi}{4} \end{align}

MSDG
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Your problem arises from using the fact that $f_x = 6x$ and $f_y = 6y$. Indeed this isn't the case, as $z^2 = 3(x^2+y^2)$, not $z$. Thus you would have:

$$z=f(x,y) = \sqrt{3(x^2+y^2)}$$ $$f_x = \frac{\sqrt{3}x}{\sqrt{x^2+y^2}} \quad \quad f_y = \frac{\sqrt{3}y}{\sqrt{x^2+y^2}}$$ Then we have:

$$\iint_S (x^2+y^2)\sqrt{1+f_x^2 + f_y^2}dxdy = \iint_S (x^2+y^2)\sqrt{1+\frac{3x^2+3y^2}{x^2+y^2}} dxdy = \iint_S 2(x^2+y^2) dxdy$$

Now use polar coordinates and compute the integral. But be careful that $\theta \in \left(0,\frac \pi2\right)$, as $x,y$ are positive.

Stefan4024
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