Given a vector field $\mathbf{F}(x,y,z) = (2xz,0,(x-1)^2)$ and three points $P_1 = (1,0,0)$, $P_2 = (0,1,0)$, $P_3 = (0,0,2)$ in $\mathbb{R^3}$. Let $T$ be the tetrahedron with corners in $P_1, P_2, P_3$ and $(0,0,0)$, and let $\mathbf{\hat{N}}$ be the unit normal vector on the surface $\partial T$ pointing out of $T$.
Compute $$\int \int_{\partial T} \mathbf{F} \cdot \mathbf{\hat{N}}dS.$$
My attempt: The plane passing through $P_1,P_2$ and $P_3$ is given by $2x + 2y + z = 2$. This gives $\mathbf{\hat{N}}dS = (2,2,1)dydx$, which further gives $\mathbf{F} \cdot \mathbf{\hat{N}}dS = (-7x^2 + 6x – 8xy + 1) dydx$. By integrating over the projection of $T$ in the $xy$-plane, we obtain $$\int \int_{\partial T} \mathbf{F} \cdot \mathbf{\hat{N}}dS = \int_0^1 \int_0^{1-x}(-7x^2 + 6x – 8xy) dydx = \frac{7}{12} .$$
However the correct answer is 1/3. Is anyone able to identify what I'm doing wrong? I'm aware that this can be solved using the divergence theorem, however I'm trying to compute the surface integral directly.
Best Answer
I believe that's because you've computed the surface integral only over one surface of tetrahedron; compute the integral for the other faces and sum them up, then we do obtain the correct value of $\frac{1}{3}$.
You have already computed the value of the integral for face $\ S_1$, which is $\frac{7}{12}$.
Now, observe that the given vector field F has no y-component, so we need not compute the integral of the face $\ S_3$ which has its unit normal along the $\ -j$ direction. It will be 0.
For the rest, we will have to compute the integrals. For $\ S_2$, whose normal is along $\ -k$:
F $\cdot \ n_2 \space dS = -(x-1)^2 \ dx \ dy$
The bounds of the projection of the given tetrahedron onto the $\ xy$ plane will be $\ 0 <= x <= 1$ & $\ 0 <= y <= (1-x)$.
Computing the integral, we observe the value to be $ \frac{-1}{4}$.
Similarly, the surface $\ S_4$ whose normal is along $\ -i$:
F $\cdot \ n_4 \space dS = -2xz \ dz \ dy$
The bounds of the projection of the given tetrahedron onto the $\ yz$ plane will be $\ 0 <= y <= 1$ & $\ 0 <= z <= 2(1-y)$.
Since $\ x = 0$ for $\ S_4$, the surface integral yet again becomes zero.
Thus, the required integral is $\frac{7}{12} + \frac{-1}{4} = \frac{1}{3}$