I was attempting to compute the surface integral of $\ G(x,y,z) = \ xyz $ over the triangular surface with vertices at $\ (1,0,0)$, $\ (0,2,0)$ and $\ (0,1,1)$.
Clearly, the first step is to parametrize the region. I thought of two approaches :
The first approach was choosing $\ x$ and $\ y $ as my parameters and rewriting $\ z $ as a function of $\ x$ and $\ y$, then projecting the above triangular surface onto the $\ xy $ plane to find the upper and lower bounds of the required surface integral.
The surface is given by the equation:
$$\ 2x + y + z = 2 $$
Then, we can use the formula:
$$ \ dS = \sqrt(1+f_x^2+f_y^2) \cdot dx dy $$
Since the normal vector of our projection is along the positive z direction.
So, I expressed $\ z$ in terms of $\ x$ and $\ y$, substituted $\ z(x,y) $ into G and obtained the double integral in terms of $\ x$ and $\ y$.
But I am having some trouble identifying the bounds of the double integral. It is obvious that 0 <= $\ x $ <= 1, but what should be the bounds for y?
I tried sketching the shadow of the triangular region on the $\ xy $ plane by moving the point $\ (0,1,1)$ to $\ (0,1,0)$ and computing the bounds for $\ y $ for a given $\ x$.
Doing that, I got $\ (1-x) <= y <= (2-2x) $ but I am unable to obtain the correct answer via this method.
So then, I thought of another method. I thought of doing the same thing with the $\ x-z $ plane instead of the $\ xy $ plane. The triangular surface remains the same and the integrand also remains the same if we simply swap out $\ y$ with $\ z$.
But I am also confused about the bounds in this case. If I project the point $\ (0,1,1)$ onto $\ (0,0,1)$ and the point $\ (0,2,0)$ to $\ (0,0,0)$, I get the bounds of z as $\ 0$ <= $\ z$ <= $\ (1-x) $. I am yet again getting an incorrect answer with this.
It is possible that the source I am checking for answers is incorrect itself. The answers I am obtaining via both methods is $\frac{\sqrt6}{30}$, while the answer given is $\frac{\sqrt6}{15}$.
Can someone please clarify the appropriate procedure to find the correct bounds in this question, and the correct answer if possible?
Best Answer
Project the triangle on the $y-z$ plane. Then the surface over which we're integrating is $\{(0,0,0),(0,1,1),(0,2,0)\}$, or in the $y-z$ plane, the triangle is $\{(0,0),(1,1),(2,0)\}$. Let's denote it with $D$. We have $x = (2 - y - z)/2$. If the surface is given by $(f(y,z),y,z)$, $$ dS = \sqrt{1 + f_y^2 + f_z^2}dydz = \sqrt{1 + 1/4 + 1/4}dydz = \frac{\sqrt{6}}{2}dydz. $$ So the integral is $$ \int_D yz\frac{2 - y - z}{2} \frac{\sqrt{6}}{2}dydz = \frac{\sqrt{6}}{2} \left( \int_0^1dy\left(\int_0^y yz\frac{2-y-z}{2}dz\right) + \int_1^2dy\left(\int_0^{2-y}yz\frac{2-y-z}{2}dz\right) \right) $$ $$ = \frac{\sqrt{6}}{2}\int_0^1\left(\frac{y^3}{2} - \frac{y^4}{4} - \frac{y^4}{6}\right)dy + \frac{\sqrt{6}}{2}\int_1^2\left( - \frac{y \left(2 - y\right)^{3}}{6} + \left(2 - y\right)^{2} \left(- \frac{y^{2}}{4} + \frac{y}{2}\right)\right)dy $$ $$ = \frac{\sqrt{6}}{48} + \frac{\sqrt{6}}{80} = \frac{\sqrt{6}}{30}. $$ I used
Sympyso hopefully I did not mess the details up.