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The question is: Find the surface area of the part of the sphere $x^2+y^2+z^2=4$ that lies above the plane $z=1$. I got $4\pi(\sqrt3-1)$ but the answer key says $4\pi(\sqrt2-1)$. Am I doing something wrong or is the answer key wrong?

Thanks in advance.

Edit: This is my work thus far:

We can write the surface in terms of z as $z = \sqrt{4-x^2-y^2}$ (where we take the positive root since $z\geq1 \rightarrow z\geq0$), which can then be parameterized to $\mathbf r(x,y) = <x,y,\sqrt{4-x^2-y^2}>$ with the condition that $x^2+y^2\leq3$ (because $z$ ranges from 1 to 2).

$\mathbf r_x = <1,0,\frac{-x}{\sqrt{4-x^2-y^2}}>$

$\mathbf r_y = <0,1,\frac{-y}{\sqrt{4-x^2-y^2}}>$

$\mathbf r_x \times \mathbf r_y = <\frac{x}{\sqrt{4-x^2-y^2}},\frac{y}{\sqrt{4-x^2-y^2}},1>$

and $\| \mathbf{\mathbf r_x \times \mathbf r_y} \| = \frac{2}{\sqrt{4-x^2-y^2}}$

So the surface area $A$ equals:

$A = \iint_S1dS = \iint_R\frac{2}{\sqrt{4-x^2-y^2}}dA = \int_0^{2\pi} \int_0^{\sqrt3}\frac{2r}{\sqrt{4-r^2}}drd\theta$

$= \int_0^{2\pi}[-2\sqrt{4-r^2}]_1^\sqrt3d\theta$

$= \int_0^{2\pi}2(\sqrt3-1)d\theta$

= $4\pi(\sqrt3-1)$

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    $\begingroup$ can you show what you did? $\endgroup$ – enzotib Jan 11 '15 at 19:00
  • $\begingroup$ Looks good to me right up until the last = sign. Based on the formula for the surface area of a spherical cap, I think the correct answer is simply $4\pi$. $\endgroup$ – Mark Dickinson Jan 11 '15 at 19:48
  • $\begingroup$ I added more work after the last = sign...did I make a mistake there? $\endgroup$ – Gabriel Jan 11 '15 at 20:27
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    $\begingroup$ It looks like your lower limit for the inner integral changed from $0$ to $1$. $\endgroup$ – Mark Dickinson Jan 11 '15 at 20:53
  • $\begingroup$ so then the answer is indeed $4\pi$ and the answer key was wrong...thanks! $\endgroup$ – Gabriel Jan 11 '15 at 21:17

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