Compute \(\iint_{\partial Q}\ F\bullet ndS\) using the easiest method available where \(Q\) is bounded by \(z=4-x^2-y^2, z=1\) and \(z=0\). \(F=<z^3,x^2y,y^2z>\).
This is the work I did:
Divergence of F: \)0+x^2+y^2\\
\(\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{1}r^3dz\ dr\ d\theta\\\)
\(\displaystyle =\int_{0}^{2\pi}\int_{0}^{2}r^3dr\ d\theta\\ \)
\(\displaystyle = \int_{0}^{2\pi}4d\theta\\ \)
\(\displaystyle = 8\pi \)
This is wrong. My bounds on r are wrong and I need to integrate r first, not z. My bounds on r should be: \(0\leq r\leq \sqrt{4-z}\).
I was told that if I integrat Z first, I need two integrals. Why would I need two?
This is the work I did:
Divergence of F: \)0+x^2+y^2\\
\(\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{1}r^3dz\ dr\ d\theta\\\)
\(\displaystyle =\int_{0}^{2\pi}\int_{0}^{2}r^3dr\ d\theta\\ \)
\(\displaystyle = \int_{0}^{2\pi}4d\theta\\ \)
\(\displaystyle = 8\pi \)
This is wrong. My bounds on r are wrong and I need to integrate r first, not z. My bounds on r should be: \(0\leq r\leq \sqrt{4-z}\).
I was told that if I integrat Z first, I need two integrals. Why would I need two?