Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$ 
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!

$V=4\displaystyle\int_{0}^{\pi/2} \displaystyle\int_{0}^{a\cos\theta}r\sqrt{{a^2}-r^2}\ drd\theta$
$=\dfrac{4a^3}{3}\displaystyle\int_{0}^{\pi/2} (1-\sin^3\theta)d\theta$
$=\dfrac{4a^3}{3}\left(\dfrac{\pi}{2}-\dfrac{2}{3}\right)=\dfrac{2a^3}{3}\left({\pi}-\dfrac{4}{3}\right)$
note: the factor of $4$ accounts for symmentry along $z$ -axis(into & outside page) and $y$-axis.see fig. below. $\theta$ is Polar angle taken from horizontal x-axis 

