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 = $dxdy = \pi\sqrt {2}$


$$\iint_S ydS = \sqrt {6}\pi $$

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...



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.

  • $\begingroup$ Thanks. I have problems (when solving surface integrals) with the argument of them. Let's say we had for instance $\iint_S xydS$ or $\iint_S y^2dS$. I wouldn't know how to proceed. May you please provide either a quick explanation on how to deal with this or a link? $\endgroup$
    – JD_PM
    Mar 20 '19 at 16:08
  • 2
    $\begingroup$ We have that $\iint_S f(x,y)dS =\sqrt{3} \int_A f(x,y)dxdy$ then let $X=x$, $Y=(y-1)/\sqrt{2}$ and use polar coordinates. Note that if $f(x,y)$ is $y^2$ or $xy$ the integrals are related with the moments of inertia of the ellipse. $\endgroup$
    – Robert Z
    Mar 20 '19 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.