The space surface(in yellow) $ x^2+y^2 = 2az\ $ is cut by a cylinder(in green) $x^2+y^2=3a^2 (a>0)$
How to calculate the cut out part area $A$?
I think the part is between the two planes $z=0$ and $ z= \frac32a $, and I only have the knowledge of using a formula $ \iint\limits_D \sqrt{1 + z^2_x + z^2_y} $ to calculate the $A$, and I should use it because there is an exercise for using this formula. In this case, $ z=\frac1{2a}(x^2+y^2), D: x^2+y^2 \le 3a^2 $ , so $$z_x = \frac1ax,\ z_y= \frac1ay$$ $$ A=\iint\limits_D \sqrt{1 + \frac{x^2}{a^2} + \frac{y^2}{a^2}} dxdy= \frac1a \iint\limits_D \sqrt{a^2+x^2+y^2} dxdy $$ using polar coordinates to do the remain steps: $$\frac1a \int_0^{2\pi}\int_0^{\sqrt3a} \sqrt{a^2+r^2}r dr=\frac\pi a\frac23 ({a^2+r^2})^\frac32\bigg|_0^{\sqrt3a}=\frac{14} 3\pi a^2$$

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  • $\begingroup$ In the integral the fractions should be $x^2/a^2$ and $y^2/a^2$. $\endgroup$ – Julián Aguirre Mar 1 '16 at 16:27
  • $\begingroup$ Thanks for your comment, I've correct it. $\endgroup$ – canoe Mar 2 '16 at 0:22

Your work is corrrect.

Note that you can find the surface also as the surface of revolution of the parabola $$ x^2=2az $$ around the $z-$ axis for $0\le x\le a\sqrt{3}$. This is a bit easier and gives the integral:

$$ A=2 \pi\int_0^{a\sqrt{3}}x\sqrt{1+\frac{x^2}{a^2}}dx $$

that can be solved with the substitution $1+\frac{x^2}{a^2}=t $ and gives the same result as your.


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