Using Divergence theorem to calculate flux

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Let W be the region bounded by the cylinder x2+y2=4, the plane z=x+1, and the xy-plane. Use the Divergence Theorem to compute the flux of F=z,x,y+z2 through the boundary of W.

So far I’ve gotten to the point of computing div(F) and integrating from 0 to x+1 to obtain

My problem is finding the bounds of the domain which is the circle of radius 2 centered at the origin. I understand I must use polar coordinates but since the circle is cut off by the line x=-1 I’m having trouble figuring out what the bounds for the radius should be. I think \theta goes from 2\pi/3 to 4\pi/3 (somewhat guessing the bound for theta when the radius is cut off by the line x = -1)

Answer

The divergence theorem allows us to write

\iint_S\vec{F}\cdot d\vec{S}=\iiint_W \nabla\cdot \vec{F}\; dV = \iiint_W 2z\; dV

where W is the region bounded by x^2+y^2=4, z=x+1 and z=0, i.e.,

W=\{(x,y,z)\;|\; -1\le x \le 2,\; 0\le z\le 1+x,\; -\sqrt{4-x^2}\le y \le \sqrt{4-x^2} \}

It follows that

\iiint_W 2z\; dV = \int_{-1}^2\int_0^{1+x}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}2z\; dydzdx = \int_{-1}^2\int_0^{1+x} 4z\sqrt{4-x^2}\; dzdx \\= \int_{-1}^22(1+x)^2\sqrt{4-x^2}\; dx = \frac{16\pi}{3}+\frac{9\sqrt{3}}{2}

Attribution
Source : Link , Question Author : Hendrix , Answer Author : Kuifje

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