The given rectangular equations are
$$x^2+y^2+z^2=64$$
$$(x-4)^2+y^2=16$$
Converting to cylindrical coordinates I get
$$r^2+z^2=64$$
$$r=4cos(\theta)$$
So my triple integral is
$$4\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\int_0^{\sqrt{16-r^2}}rdzdrd\theta$$
It's a long and tedious calculation but I worked through it and got $-\frac{1408\pi}{15}$ which doesn't make sense because you can't have a negative volume. The negative came from the second integral which is
$$4\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}r\sqrt{16-r^2}drd\theta$$
where I used a u-sub, where $u=16-r^2$ and $du=-2r$ and moving that $-\frac{1}{2}$to the outside gives me
$$-2\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\sqrt{u}dud\theta$$
So could I just swap the limits of integration to get rid of it or have I messed up somewhere else?
Use cylindrical coordinates to find the volume of the solid using triple integrals
cylindrical coordinatesmultivariable-calculus
Related Solutions
By cylindrical coordinates the set up of the integral should be
$$\int_0^{2\pi}\int_{-3}^{-\frac3{\sqrt2}}\int_{0}^{\sqrt{9-z^2}} r\,\, dzdrd\theta+\int_0^{2\pi}\int_{-\frac3{\sqrt2}}^0\int_{0}^{-z} r\,\, dzdrd\theta$$
In order to better understand the solid region $B$ enclosed by both surfaces in order to better evaluate our limits of integration, we can visualize the surfaces as follows:
In cylindrical coordinates, we can express the cylinder as: $$ r = a \cos \theta \qquad (0 \leq \theta \leq \pi)$$ and the sphere as: $$ r^2 + z^2 = a^2$$
We see that, as we traverse all the $(r,\theta)$ or $(x,y)$ points within the cylinder's projection on the $x$-$y$ plane (a circle), the values of $z$ we want are between $\pm\sqrt{a^2-r^2}$
Finally, we recall that: $$ dV = dx \, dy \, dz = r \, dr \, d\theta \, dz $$
So: \begin{align*} \iiint_B \, dV &= \int_{0}^{\pi} \int_0^{a \cos \theta}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} r \, dz \, dr \, d\theta \\ &= \int_{0}^{\pi} \int_0^{a \cos \theta} 2\int_{0}^{\sqrt{a^2-r^2}} r \, dz \, dr \, d\theta \\ &= \int_{0}^{\pi} \int_0^{a \cos \theta} 2r\sqrt{a^2-r^2} \, dr \, d\theta \\ &= \int_{0}^{\pi} \left[ -\frac{2}{3}(a^2-r^2)^{3/2} \right]_0^{a \cos \theta} \, d\theta \\ &= \frac{2}{3}\int_{0}^{\pi} (a^3 - a^3 \sin^3 \theta) \, d\theta \\ &= \frac{2a^3}{3}\int_{0}^{\pi} (1 - \sin^3 \theta) \, d\theta \\ &= \frac{2a^3}{12}\int_{0}^{\pi} (4 - 3\sin \theta + \sin 3\theta) \, d\theta \\ &= \frac{a^3}{6} \left[4\theta + 3 \cos \theta - \frac{1}{3}\cos 3\theta \right]_0^{\pi} \\ &= \frac{a^3}{6} \left(4\pi - 3 + \frac{1}{3} - 3 + \frac{1}{3} \right) \\ &= \frac{a^3}{6} \left(4\pi - \frac{16}{3} \right) \\ &= \frac{2a^3}{9} \left(3\pi - 4 \right) \end{align*}
The problem that you probably ran into is parametrizing from $\theta = -\pi/2$ to $\theta = \pi/2$. This is a valid parametrization, but when you reach this point: $$ L = \frac{2}{3}\int_{-\pi/2}^{\pi/2} \left[a^3 - (\sin^2 \theta)^{3/2} \right] \, d\theta $$
We have that: $$ (\sin^2 \theta)^{3/2} \neq \sin^3 \theta $$ for $\sin \theta < 0$, which occurs for $-\pi/2 \leq \theta < 0$.
So we choose a parametrization where we keep $\sin \theta \geq 0$, which is why we use $0 \leq \theta \leq \pi$.
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Best Answer
There are two mistakes I notice. You have written wrong bound of $z$ and the equation of cylinder that was already pointed out. The correct integral should be
$V = 4\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{8\cos(\theta)}\int_{0}^{\sqrt{64-r^2}}rdzdrd\theta \approx 617.22$
$V = 4\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{8\cos(\theta)} r \, \sqrt{64-r^2} \, dr \, d\theta$
On substition, $64-r^2 = u, rdr = -\frac{1}{2}du$.
The bound $r = 0$ becomes $u = 64$, $r = 8 \cos \theta$ becomes $u = 64 \sin ^2 \theta$.
$V = -2\displaystyle \int_0^{\frac{\pi}{2}}\int_{64}^{64\sin^2(\theta)} \sqrt u \, du \, d\theta = 2\int_0^{\frac{\pi}{2}}\int_{64\sin^2(\theta)}^{64} \sqrt u \, du \, d\theta$
$V = \displaystyle \frac{2048}{3} \int_0^{\frac{\pi}{2}} (1-sin^3 \theta) \, d\theta = \frac{2048}{3} (\frac{\pi}{2} - \frac{2}{3})$