Triple Integrals in Cylindrical coordinates

**aprilkravitz** · April 28th 2008, 08:01 AM

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.

**Danshader** · April 28th 2008, 09:03 AM

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} $

Thread: Triple Integrals in Cylindrical coordinates

LinkBack

Thread Tools

Display

Triple Integrals in Cylindrical coordinates

Similar Math Help Forum Discussions

Using cylindrical coordinates to solve triple integrals

Cylindrical Coordinates Triple Integral

Triple Integrals In Cylindrical Coordinates.

Triple integral with cylindrical coordinates

cylindrical coordinates = triple integral

Search Tags