Question

ques

Answer 1

I have to find
\[\iint_\limits S \vec A.\hat n ds\]
Where
\[\vec A=z \hat i+x \hat j-3y^2z \hat k\]
For this I've divided the surface into 3 regions, s1 s2 and s3

So we have
\[\iint_\limits S \vec A . \hat n ds =\iint_{S_{1}} \vec A . \hat n ds+ \iint_{S_{2}}\vec A. \hat n ds+\iint_{S_{3}}\vec A. \hat nds\]
For S1 we have
\[\hat n=-\hat k\](outward normal)
\[z=0\]
and
\[ds=dxdy\]
\[\vec A . (-\hat k)=3y^2z\]
So
\[\iint_{S_{1}}\vec A . \hat n ds=\iint_{S_{1}}3y^2zdxdy=0\]
because z is 0 in the xy plane
For S3 we have
\[\hat n=\hat k\]
\[z=5\]
\[ds=dxdy\]
\[\vec A . \hat k=-3y^2z=-15y^2\]
\[\iint_{S_{3}} \vec A . \hat n ds=\iint_{S_{3}} -15y^2dxdy=\int\limits_{0}^{4}\int\limits_{0}^{\sqrt{16-x^2}}-15y^2dydx=-5\int\limits_{0}^{4}(16-x^2)^\frac{3}{2}dx\]
\[\iint_{S_{3}}\vec A.\hat n ds=-240\pi\]
For S2 I considered it's projection on xz plane
\[\iint_{S_{2}}\vec A .\hat n ds=\iint_\limits R \vec A.\hat n \frac{dxdz}{|(-\hat j).\hat n|}\]
Here
\[\hat n=\frac{x \hat i+y \hat j}{\sqrt{x^2+y^2}}=\frac{x \hat i+y \hat j}{4}\]
so we have
\[\vec A. \hat n=\frac{xz}{4}+\frac{xy}{4}=\frac{xz}{4}\]
(y is 0 in xz plane)
and \[ds=\frac{dxdy}{|(-\hat j).\hat n|}=\frac{\sqrt{x^2+y^2}}{y}dxdz=\frac{4}{y}dxdz\]
\[\iint_\limits R \vec A . \hat n ds=\int\limits_{0}^{5}\int\limits_{0}^{4}\frac{xz}{y}dxdz\]
Now since y becomes 0 this is kind of undefined and I'm stuck here :S