integrationvolume
I'm having problems with computing the volume of the solid bounded by the cone $z = 3\sqrt{x^2 + y^2}$, the plane $z = 0$, and the cylinder $x^2 + (y-1)^2 = 1$.
I tried substituting cylindrical coordinates and computed $r = 2sin\theta$, then I integrated $\int_0^{2\pi}\int_0^{2sin\theta}\int_0^{3r} \ \text{r dz dr d} \ \theta$, but I ended up getting 0 for some reason. The correct answer is $\frac{32}{3}$. Anyone knows what went wrong and how I should go about solving this question?
Thanks!
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
$$V= \left (2 \pi - \frac{32}{9}\right )a^3$$
The formula
$$V=\int_?^?\int_?^?\int_?^? r\> dz\>dr\>d\theta$$
refers to cylindrical coordinates. In particular, $\theta$ is the polar angle centered at $(0,0)$. Even though $\theta$ is not the "natural" parameter for the circle $(x-3)^2+y^2=9$ you can use $\theta$ to describe this circle in terms of $r$ and $\theta$ as well. Looking at the figure below one immediately verifies that the circle has the polar representation
$$r(\theta)=6\cos\theta\qquad\left(-{\pi\over2}\leq\theta\leq{\pi\over2}\right)\ .$$
The volume in question is then given by
$$V=\int_{-\pi/2}^{\pi/2}\int_0^{6\cos\theta}\int_0^r r\> dz\>dr\>d\theta=96\ .$$
Best Answer
The parameterized (polar) curve $$r(\theta) = 2 \sin \theta$$ can be written as the parameterized (Cartesian) curve $$(x(\theta), y(\theta)) = (2 \sin \theta \cos \theta, 2 \sin^2 \theta) = (\sin 2 \theta, \cos 2 \theta + 1);$$ since both of the components in the last expression have period $\pi$, the parameterization traces out the circular boundary twice over the interval $[0, 2 \pi]$; we only want to integrate over each direction $\theta$ once, so we should replace the interval over which we integrate $\theta$ by $[0, \pi]$.
We can see too why the triple integral you wrote evaluates to zero: The upper limit on the middle integral, is $2 \sin \theta$, and the symmetry of that function (together with the fact that $\theta$ appears nowhere else in the integral) ensures that the integral over $[\pi, 2\pi]$ has the same magnitude but the opposite sign as the desired integral over $[0, \pi]$.