Find volume inside the cone $z= 2a-\sqrt{x^2+y^2}$ and inside the cylinder $x^2+y^2=2ay$

I have this question and I have seen here that there are similar questions but I have been trying to figure it out for a while with no luck.

I have to find volume inside the cone $z= 2a-\sqrt{x^2+y^2}$ and inside the cylinder $x^2+y^2=2ay$.

I am trying to use cylindrical coordinates: $x=r\cos\theta\,\,\, y=r\sin\theta\,,\, z=z$.

I think I am mainly having trouble finding the boundaries for the integration $\iiint rdrdzd\theta$:

$0\leq \theta\leq 2\pi$, $0\leq z \leq 2a$, and for $r$. I think this is where I am wrong... I have $0 \leq r \leq z\,\sin(\theta)/(1-\sin(\theta)).$

If anyone knows or can give me some help I would appreciate it.

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What do you mean by ZZZ in "the integration ZZZ $rdrdzd\theta$"? Is it $\iiint rdrdzd\theta$? –  Américo Tavares Mar 25 '13 at 13:42
Fixed TeX code for $\iiint rdrdzd\theta$. –  Américo Tavares Apr 21 '13 at 11:25

The cross-sectional area of this volume is the intersection between the circular cross-sections of the cone and cylinder. If we call that cross-sectional area $A(z)$, then the volume is

$$V = \int_0^{2 a} dz \: A(z)$$

So the trick is to find $A(z)$. There's no substitute in drawing a picture:

In red is the cylinder cross-section, and in blue are cone cross-sections at various values of $z$. The integration limits will depend on $z$: when the cone cross-section intersects the cylinder cross-section below the center of the cylinder cross-section, then the intersection is bounded by the upper arc of the cone and the lower arc of the cylinder. When the cone CS intersects the cylinder CS above the center of the cylinder CS, however, there is also additional area to the left and right of the intersection points.

The intersection points are easy enough to find; they are, in polars:

$$\sin{\theta} = 1-\frac{z}{2 a}$$

The cross sectional area at $z$ is the sum of two contributions: one ($A_1$) bounded by the red circle, the other ($A_2$) by the blue circle.

$$A_1 = \int_{ \arcsin{(1-z/(2 a))}}^{\pi - \arcsin{(1-z/(2 a))}} d\theta \: \int_0^{2 a-z} dr \, r = (2 a-z)^2 \left[\pi - 2\arcsin{\left(1-\frac{z}{2 a} \right)}\right]$$

$$A_2 = 2 \int_0^{\arcsin{(1-z/(2 a))}} d\theta \: \int_0^{2 a \sin{\theta}} dr \, r = 2 a^2 \left [ \arcsin{\left(1-\frac{z}{2 a} \right)} - \left(1-\frac{z}{2 a} \right) \sqrt{1-\left(1-\frac{z}{2 a} \right)^2} \right ]$$

The cross-sectional area $A(z)=A_1+A_2$; we integrate over $z$ to get the volume. We may make the substitution $u=1-z/(2 a)$ and get

$$V=4 a^3 \int_0^1 du \: [\pi u^2 +(1- 2 u^2) \arcsin{\sqrt{1-u^2}} -u \sqrt{1-u^2}]$$

or

$$V= \left (2 \pi - \frac{32}{9}\right )a^3$$

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Thanks for the hgelp –  user68203 Mar 24 '13 at 3:22
I really appreciate the help however I am a bit confused with your expression for A(z). I see how you got x and y however how did you go from there to the expression for A(z)? Also I am working on changing of variables in class so how can a apply what you said, change the variables, then solve the question? –  user68203 Mar 24 '13 at 3:29
$A$ is the intersection of the circles. I find it easiest to work in Cartesians to set up, and then if you want to change variables, that's OK. The expression for $A$ is complicated because the behavior of the intersection through $z$ is complicated. It's difficult to express how I went from the intersections to the expression, but really, use the picture as your guide. –  Ron Gordon Mar 24 '13 at 3:44
This is a bit easier for me to understand however the answer is not the answer in the back of the book so I am still a bit confused. The answer is supposed to be (2pi-32/9)a^3... Is there not a way to use cylindrical coordinates where I just find the boundaries then do a simple triple integration. It seems to me that the boundaries for theta would be 0<theta<pi, then I am just a bit confused with what the boundaries for r and z would then be... –  user68203 Mar 24 '13 at 16:16
I have tried a new route but seem to be stuck again. If you wouldn't mind looking it over to see if I am right or if my calculations are off I would appreciate it. So far I now have ZZZ rdrd(theta)dz with boundaries 0<z<2a-r, 0<r<asin(theta), 0<theta<2pi –  user68203 Mar 25 '13 at 2:09

Converting the Cartesian equation of the base ($z=0$) of the given vertical cylinder, which is centered at $(0,a)$ and has radius $a$

$$x^{2}+y^{2}=2ay\Leftrightarrow x^{2}+\left( y-a\right) ^{2}=a^{2}\tag{1}$$

to polar coordinates $x=r\cos \theta ,y=r\sin \theta$ I got

$$r=2a\sin \theta ,\tag{1a}$$

because $x^{2}+y^{2}=r^{2}$, $r\ge 0$, and $$\begin{eqnarray*} x^{2}+y^{2}-2ay &=&0\Leftrightarrow r^{2}-2ar\sin \theta =0 \\ \Leftrightarrow r\left( r-2a\sin \theta \right) &=&0\Leftrightarrow r=2a\sin \theta .\tag{1b} \end{eqnarray*}$$ Note that $(1\mathrm{a})$ may be found directly from the figure below using simple trigonometry.

$$\text{Disk }R=\{(x,y)\in\mathbb{R}^2: 0\le x^{2}+\left( y-a\right) ^{2}\le a^{2} \};\, 0\leq r\leq 2a\sin \theta , 0\leq \theta \leq \pi .$$

The volume of the region inside the cylinder and bounded from below by the plane $z=0$ and from above by the given cone

$$z=2a-\sqrt{x^{2}+y^{2}}=2a-r\tag{2}$$

can be expressed by the double integral $$V=\iint_{R}2a-\sqrt{x^{2}+y^{2}}\,dA,\tag{3}$$ where $R$ is the disk whose boundary is $(1)$. In polar coordinates $dA=dx\, dy=r\,dr\,d\theta$ and $R$ is defined by $0\leq r\leq 2a\sin \theta$, with $0\leq \theta \leq \pi$. So $(3)$ is transformed into $$\begin{eqnarray*} V &=&\int_{0}^{\pi }\left( \int_{0}^{2a\sin \theta }\left( 2a-r\right) \,r\,dr\right) \,d\theta \tag{4} \\ &=&\int_{0}^{\pi }\left. ar^{2}-\frac{1}{3}r^{3}\right\vert _{0}^{2a\sin \theta }\,d\theta \\ &=&a\int_{0}^{\pi }\left( 2a\sin \theta \right) ^{2}\,d\theta -\frac{1}{3} \int_{0}^{\pi }\left( 2a\sin \theta \right) ^{3}\,d\theta \\ &=&2a^{3}\pi -\frac{32}{9}a^{3}=\left( 2\pi -\frac{32}{9}\right) a^{3}, \end{eqnarray*}$$ where the integrals of $\sin^2\theta$ and $\sin^3\theta$ were evaluated as follows: $$\begin{eqnarray*} \int_{0}^{\pi }\sin ^{2}\theta \,d\theta &=&\int_{0}^{\pi }\frac{1-\cos 2\theta }{2}\,d\theta =\frac{\pi }{2},\tag{4a} \\ \int_{0}^{\pi }\sin ^{3}\theta \,d\theta &=&\int_{0}^{\pi }\frac{3\sin \theta -\sin 3\theta }{4}\,d\theta =\frac{4}{3}.\tag{4b} \end{eqnarray*}$$

Final comment. Perhaps in a closer relation to the spirit of the question the double integrals $(3)$ and $(4)$ may be written as triple integrals, respectively,

$$\begin{equation*} V=\iint_{R}\left( \int_{0}^{2a-\sqrt{x^{2}+y^{2}}}dz\right) \,dA \end{equation*}\tag{3^\prime}$$ and $$\begin{equation*} V=\int_{0}^{\pi }\left( \int_{0}^{2a\sin \theta }\left( \int_{0}^{2a-r}dz\right) \,r\,dr\right) \,d\theta .\tag{4^\prime} \end{equation*}$$

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