# Thread: Triple Integrals in Cylindrical coordinates

1. ## Triple Integrals in Cylindrical coordinates

Find the volume of the region above the plane z=1/2 and below the sphere x^2+y^2+z^2=1 with a triple integral in cylindrical coordinates.

2. well from the question given this what i can think of:
$
V = \int\int\int dV
$

$
V = \int^{1}_{-1} \int ^{\sqrt{1-x^2}} _ {-\sqrt{1-x^2}} \int ^{\sqrt{1-x^2 - y^2}} _{\frac{1}{2}} dzdydx$
------- rectangular form

rules of polar coordinates:
$r^2 = x^2 + y^2
$

$x = r.cos\theta
$

$y=r.sin\theta
$

$
V=\int^{2 \pi}_{0} \int ^{1} _ {0} \int ^{\sqrt{1-r^2}} _{\frac{1}{2}} r.dzdrd\theta$
--------------cylindrical form
$
= \int^{2 \pi}_{0} \int ^{1} _ {0} r.z |^{\sqrt{1-r^2}} _\frac{1}{2}.drd\theta
$

$
= \int^{2 \pi}_{0} \int ^{1} _ {0} r.(\sqrt{1-r^2} - \frac{1}{2}).drd\theta
$

$
= \int^{2 \pi}_{0} \int ^{1} _ {0} r\sqrt{1-r^2} - \frac{r}{2}.drd\theta
$

$
=\int^{2 \pi}_{0} -\frac{1}{3} {(1-r^2)}^{3/2}|^1 _0 - \frac{r^2}{4}|^1 _0.d\theta
$

$
=\int^{2 \pi}_{0} \frac{1}{12}.d\theta
$

$
=\frac{2\pi}{12}
$

$
=\frac{\pi}{6}
$