# Area of the part of the cylinder $x^2+y^2=2ay$ outside the cone $z^2=x^2+y^2$

Problem: Find the area of the part of the cylinder $$x^2+y^2=2ay$$ that lies outside the cone $$z^2=x^2+y^2$$.

My attempt: So I thought we could do this by projecting the surface onto the $$yz$$-plane and taking the surface integral of the function $$x=g(y,z)=\sqrt{z^2-y^2}$$. I.e letting $$S$$ be the surface and $$E$$ be the projection onto the $$yz$$-plane where we have a $$2$$ before the integral over $$E$$ since we have both $$x<0$$ and $$0\leq x$$: \begin{align*}\iint_{\mathcal{S}}x \ \mathrm{d}S &=2\iint_{E}x\underbrace{\sqrt{1+\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2} \ \mathrm{d}z\mathrm{d}y}_{\mathrm{d}S} \\ &=2\iint_{E}x\sqrt{1+\frac{z^2}{x^2}+\frac{y^2}{x^2}} \ \mathrm{d}z\mathrm{d}y\\ &=2\iint_{E}\sqrt{x^2+z^2+y^2}\ \mathrm{d}z\mathrm{d}y\\ &=2\iint_{E}\sqrt{2}z\ \mathrm{d}z\mathrm{d}y \end{align*} Now in the projection it seems to me that we have the following bounds on $$z$$ and $$y$$ since the cylinder has radius $$a$$ and the cone and the surface intersect at $$z=\sqrt{2ay}$$ $$0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$$ so: \begin{align*}2\iint_{E}\sqrt{2}z\ \mathrm{d}z\mathrm{d}y &= \sqrt{2}\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}2z \ \mathrm{d}z\mathrm{d}y \\ &=\sqrt{2}\int_{0}^{2a} 2ay \ \mathrm{d}y\\ &=4\sqrt{2}a^{2}\end{align*} However my book says its $$16a^2$$ so what is my mistake(s)?

PS. I think this is also possible with polar coordinates but I would like to use the surface integral with projection onto the $$yz$$-plane.

PSDS. Picture is not totally acurate as $$a=4$$

Edit:

As Ninad Munshi pointed out I was projecting the wrong surface and I used the wrong formula for the surface area. My thoughts are

Would it be correct to say that $$\iint\mathrm{d}S$$ is the surface area, and would $$\mathrm{d}S$$ be $$\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2} dzdy$$? If so I still seem to be off by a factor of $$2$$ as \begin{align*}\iint_{\mathcal{S}} \mathrm{d}S &= 2 \iint_{E}\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2}dzdy \\ &=2\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2}dzdy=8a^{2}\end{align*}

• You used the wrong surface in the beginning. It wants the area of the cylinder, not the cone, so you have to project the cylinder down. – Ninad Munshi May 12 '20 at 21:33
• Also, $\iint x\:dS$ does not give you surface area. – Ninad Munshi May 12 '20 at 21:35
• @NinadMunshi I solved it, Thank you for your help! – André Armatowski May 13 '20 at 2:42

The correct way to solve this question is to start with the cylinder $$x^2+y^2=2ay$$ that we wish to project onto the $$yz$$ plane. This is done by first calculating $$\mathrm{d}S$$ in $$\iint_{\mathcal{S}}dS$$ which gives the surface area
We have that $$dS = \sqrt{1+\left(\frac{\partial x}{\partial y}\right)^{2}+\left(\frac{\partial x}{\partial z}\right)^{2}} \ \mathrm{d}z\mathrm{d}y=\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$
Now the problem I had is that when I projected the cylinder: I only considered the symmetric areas of $$x<0$$ and $$0\leq x$$ while we infact have two more symmetries: namely $$z<0$$ and $$0\leq z$$.
In summary: We have four areas that are equal (and not two) so letting $$E$$ represent the area of the projection of the cylinder onto the $$yz$$-plane in the first octant we get: $$\iint_{\mathcal{S}}\mathrm{d}S=4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$ The particular limits of $$z$$ and $$y$$ are still correct that is $$0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$$ So: \begin{align*}4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y & = 4\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y \\ &= 4 \int_{0}^{2a}\sqrt{2ay}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}y\\ &= 4(4a^2)=16a^2\end{align*} Which is the correct answer