I'm trying to solve this but my answer was wrong every time. The question is:
If $\vec{F} = 2y\hat{i} – 3\hat{\jmath} + x^2\hat{k}$ and $S$ is the
surface of the parabolic cylinder $y^2 = 8x$ in the first octant
bounded by the planes $y = 4$ and $z = 6$, then evaluate
$$\iint_S\vec{F}\cdot\hat{n}\ dS.$$
Correct answer is : 132
I've tried to integrate by taking the projection of the surface $y^2 = 8x$ in $yz$ plane, where I've taken $\hat{n} = \hat{i}$ as unit normal to $yz$ plane.
Best Answer
Parametrize your surface, say by:
$$\vec r(u,v):=\left(\frac{u^2}8\,,\,\,u\,,\,\,v\right)\;,\;\;0\le u\le 4\;,\;\;0\le v\le 6\;\implies$$
$$\vec{r'_u}=\left(\frac u4\,,\,\,1\,,\,\,0\right)\;,\;\;\vec{r_v'}=(0,0,1)\implies\vec{r'_y}\times\vec{r'_v}=\begin{vmatrix}\vec i&\vec j&\vec k\\\frac u4&1&0\\0&0&1\end{vmatrix}=\left(1,\,-\frac u4\,,\,\,0\right)$$
Then, since $\;\vec n=\frac{\vec{r'_y}\times\vec{r'_v}}{\left\|\vec{r'_y}\times\vec{r'_v}\right\|}\;$ ( and observe that the norm of the vectorial product cancels in the following!) and also $\; F(\vec r(u,v))=\left(2u\,,\,\,-3\,,\,\,\frac{u^4}{64}\right)\;$ , we get that:
$$\iint_s\vec F\cdot\vec n\;d\vec S=\iint_D\vec F(\vec r(u,v))\cdot\left(\vec{r'_y}\times\vec{r'_v}\right) dA=\int_0^4\int_0^6\left(2u\,,\,-3\,,\,\frac{u^4}{64}\right)\cdot\left(1\,,\,-\frac u4\,,\,\,0\right)\,dA=$$
$$=\int_0^4\int_0^6\frac{11}4u\,dv\,du=\frac{66}4\int_0^4u\,du=\frac{33}4\cdot16=132$$