This is ridiculous but I don't see how to find the between two surfaces.

I'm trying to evaluate the surface integral $$\iint_{\partial S} \mathbf{F}\cdot \mathbf{n} dA$$ where $F(x,y,x)=x\hat{i}+y\hat{j}+z\hat{k}$ and S is the boundary of the region $x^2+y^2\leq x\leq \sqrt{2-x^2-y^2}$ oriented so that the normal points out of the region. I want to do it using the divergence theroem.

So basically what I am stuck on is finding the intersection of those two surfaces because from there I would have a region for my double integral in this

Question

This is ridiculous but I don't see how to find the between two surfaces.

I'm trying to evaluate the surface integral $$\iint_{\partial S} \mathbf{F}\cdot \mathbf{n} dA$$ where $F(x,y,x)=x\hat{i}+y\hat{j}+z\hat{k}$ and S is the boundary of the region $x^2+y^2\leq x\leq \sqrt{2-x^2-y^2}$ oriented so that the normal points out of the region. I want to do it using the divergence theroem.

So basically what I am stuck on is finding the intersection of those two surfaces because from there I would have a region for my double integral in this

richyw · Answer

$$\iint_R \int^{\sqrt{2-x^2-y^2}}_{x^2+y^2}3dzdA$$

richyw · Answer

I mean I hacked away at it like this 

$$(x^2+y^2)^2=2-(x^2+y^2)$$$$(x^2+y^2)(x^2+y^2+1)=2$$which makes it $x^2+y^2=1$ but honestly that's sketchy. How do I find this is a systematic test-taking no brain way?

anonymous · Answer

whitch to polar system..

richyw · Answer

how does that make the equality more obvious?

anonymous · Answer

Wow very complicated sorry I can't help.

anonymous · Answer

in polar form, your region becomes,
$$r^2\le r\cos\theta\le\sqrt{2-r^2}$$

anonymous · Answer

two circles .. one centered at origin and one shifted along x-axis