Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$
Attempt: The projection of the Cylinder ( denoted $D$) on the $xy$ plane is a circle which has the equation: $x^2+y^2=ax ~~~\equiv~~~(x-a/2)^2+y^2=(a/2)^2 ~~~\equiv~~~~r=a \cos \theta$ ( In Polar Coordinates)
The circle has centre $(a/2,0)$ and radius $a/2$. Hence, the $y$ axis is a tangent to it.
Hence, the required volume $V = \int \int \int dv $
$= \int \int_D \int_{-\sqrt {a^2-x^2-y^2}} ^{\sqrt {a^2-x^2-y^2}}dz~~dx~dy$
$=2\int \int_D \sqrt {a^2-x^2-y^2}~~ dx dy$
Switching to Polar Coordinates :
$V=2\int_{-\pi/2}^{\pi/2} \int_{0}^{a \cos \theta} \sqrt {a^2-r^2}~~ r~dr~ d\theta=\dfrac {2}{3}a^3 \pi$
Although, the limits of $\theta$ look conceptually fine to me, my textbook uses the limits $0$ to $\pi$ and gives the result $= \dfrac {2}{3}a^3 (\pi-\dfrac{4}{3})$.
Could someone please clarify why $-\pi/2$ to $\pi/2$ usage in my expression might be incorrect?
Thanks a lot for your help!
$V=2\int_{-\pi/2}^{\pi/2} \int_{0}^{a \cos \theta} \sqrt {a^2-r^2}~~ r~dr~ d\theta\\ V=2\int_{-\pi/2}^{\pi/2}-\frac 13 (a^2-r^2)^\frac 32|_0^{a\cos\theta}~~ d\theta$
When we evaluate the limit at $a\cos \theta$....
$(a^2-a^2\cos^2\theta)^\frac 32\\ (a^2\sin^2\theta)\frac 32\\ |a^3\sin^3\theta|$
That is we only consider the principal root.
Or, $a^2-a^2\cos^2\theta \ge 0$ for all $\theta$ and so the cube of the square root.
$\frac 23 \int_{-\pi/2}^{\pi/2}a^3 - |a^3\sin^3 \theta|~~ d\theta = \frac 23 \int_{0}^{\pi}a^3 - |a^3\sin^3 \theta|~~ d\theta = (\frac 23\pi - \frac 43) a^3$
And the limits of integration don't really matter.