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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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