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I started by drawing both graphs and found that the intersection is just the part of the sphere above $z=1$. So it's the part of the sphere from $1\le z\le2$ and let this be called $S^1$. I then let $g(x,y,z)=x^2+y^2+z^2$ and $f(x,y)=z=\sqrt{4-x^2-y^2}$ I then used the formula: $$ \int\int_Sg(x,y,z)\ ds=\int\int_Dg(x,y,f(x,y))*\sqrt{1+(f_x)^2+(f_y)^2}\ dA$$ Plugging in my values I got: $$ g(x,y,f(x,y))=x^2+y^2+(\sqrt{4-x^2-y^2})^2=x^2+y^2+4-x^2-y^2=4 $$ and $$ \sqrt{1+(f_x)^2+(f_y)^2}=\sqrt{1+\frac{x^2}{4-x^2-y^2}+\frac{y^2}{4-x^2-y^2}}=\frac{2}{\sqrt{4-x^2-y^2}} $$ so $$ \int\int_Dg(x,y,f(x,y))*\sqrt{1+(f_x)^2+(f_y)^2}\ dA=\int\int_D\frac{8}{\sqrt{4-x^2-y^2}} $$ I then switched to polar letting $x=r\cos\theta$ and $y=r\sin\theta$. So after pulling back the one forms and substituting I was left with: $$ =\int\int\frac{8r}{\sqrt{4-r^2}} dr\ d\theta $$ From here I'm starting to get a little confused. I know that at $z=1$ the region $z\ge1$ intersects $S$ and forms a circle in this plane. So I think that $r$ should be integrated from $0\le r\le \sqrt(3)$ -- I found $\sqrt(3)$ by letting $z=1$ and $y=0$ and then $\max(x)=\sqrt(3)$. My concern however is that in the $z$ direction still goes up to $2$ so I don't think this is correct. I do not however have any other idea how to find the interval for $r$ though. I am also equally confused about the interval that would be appropriate for $\theta$ as well.

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    $\begingroup$ Volume? Why are you doing a surface integral? $\endgroup$ – Ted Shifrin Apr 30 at 4:14
  • $\begingroup$ sorry.. the original problem is to calculate the area of the intersection $\endgroup$ – joseph Apr 30 at 4:24
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    $\begingroup$ Yes, of course. Try doing it in both cylindrical and spherical coordinates. $\endgroup$ – Ted Shifrin Apr 30 at 4:26
  • $\begingroup$ was my use of polar incorrect? $\endgroup$ – joseph Apr 30 at 4:28
  • $\begingroup$ and thank you I will edit my original post right now $\endgroup$ – joseph Apr 30 at 4:30
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What you have so far seems right at a glance. There are a few key insights that should make the rest of the problem relatively easy to solve.

The first is that polar $r$ represents an angle in the plane. So if you imagine the shadow of the region you're integrating in the $xy$-plane, it will be a circle. But what's the radius? Well, it's based on the radius of the cross-section of the sphere at $z=1$. So if $x^2+y^2+z^2=4$, then $x^2+y^2=3$, and it leads to the $r=\sqrt{3}$ that you got. And $\theta$ comes into play because you imagine $r$ going from $0$ to $\sqrt{3}$ all the way around the circle.

So where is the height taken to account? It's implicitly hidden in the function $g$, which accounts for the two surfaces you're integrating. By integrating this way, you're turning a triple integral into a double integral of surfaces. So indeed, $0\leq r\leq\sqrt{3}$.

The other reminder is that when you convert to polar, use area scaling factor $rdrd\theta$. With these facts in mind, the rest should be simple.

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  • $\begingroup$ thank you for the help! I just edited my original answer to take into account the scaling factor $\endgroup$ – joseph Apr 30 at 4:10

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