Calculating the area of the intersection between $S: x^2+y^2+z^2=4$ and $z\ge1$.

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.

• Volume? Why are you doing a surface integral? – Ted Shifrin Apr 30 at 4:14
• sorry.. the original problem is to calculate the area of the intersection – joseph Apr 30 at 4:24
• Yes, of course. Try doing it in both cylindrical and spherical coordinates. – Ted Shifrin Apr 30 at 4:26
• was my use of polar incorrect? – joseph Apr 30 at 4:28
• and thank you I will edit my original post right now – joseph Apr 30 at 4:30

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.