Find $\iint_S ydS$, where $s$ is the part of the cone $z = \sqrt{2(x^2 + y^2)}$ that lies below the plane $z = 1 + y$
The intersection of these two is an ellipse of area $A = \pi\sqrt {2}$
Note that this problem has been solved here: Surface integral problem. However, I found a slightly different approach.
Let's calculate dS:
$$dS = \sqrt{1 + \frac{4(x^2 + y^2)}{z^2}}dxdy$$
Plugging $z = \sqrt{2(x^2 + y^2)}$ into it you get:
$$dS = \sqrt{3}dxdy$$
Where A = $dxdy = \pi\sqrt {2}$
Then:
$$\iint_S ydS = \sqrt {6}\pi $$
Note this is not the same result robjohn got.
I don't understand why $y$ is treated as it wasn't there.
I understand the following: $\iint_S dS = \sqrt {6}\pi $
But we have $\iint_S ydS$ and not $\iint_S dS$. I guess there must be a symmetry argument to justify this but I don't see it…
Thanks.
Best Answer
According to your approach $$\iint_S ydS =\sqrt{3} \int_A ydxdy=\sqrt{3}\, \bar{y}|A|=\sqrt{3} |A|=\sqrt{6}\pi$$ where $A$ is the interior of the ellipse $x^2+\left(\frac{y-1}{\sqrt{2}}\right)^2=1$, $(\bar{x},\bar{y})=(0,1)$ is its centroid (which coincides with its geometric center), and $|A|=\sqrt{2}\pi$ is its area.