Evaluate the surface integral where is the part of the plane in the first octant.
A) 3
B) 192
D) 400
E) 800 $\pi$

Question

Evaluate the surface integral  where  is the part of the plane  in the first octant.
A) 3 
B) 192 
D) 400 
E) 800 $\pi$

Quizplus · Accepted Answer

We can parameterize the plane as $\vec{r}(u,v) = \begin{pmatrix}u\ v\ 3-2u-3v\end{pmatrix}$, where $0\leq u\leq 1$ and $0\leq v\leq 1$. Then the normal vector is given by $$ \vec{n} = \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v} = \begin{pmatrix}2\ 3\ 1\end{pmatrix}. $$ To ensure that the normal vector points in the positive $z$-direction, we need to orient the surface to have $\vec{n}$ pointing towards the origin. Hence the surface integral becomes $$ \iint_S \begin{pmatrix}1\ 1\ 1\end{pmatrix}\cdot \begin{pmatrix}2\ 3\ 1\end{pmatrix} dS = \int_0^1\int_0^1 \begin{pmatrix}2\ 3\ 1\end{pmatrix}\cdot \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v} du dv = \int_0^1\int_0^1 11\,du\,dv = \boxed{\textbf{(C)}\ 11}. $$

Evaluate the Surface Integral Where Is the Part π\piπ

Evaluate the Surface Integral Where Is the Part $\pi$