Triple Coordinates

**tdat1979** · November 2nd 2009, 06:15 AM

Use spherical coordinates to find the volume of the region above the cone z = $\sqrt {x^2 + y^2}$ and between the hemispheres z = $\sqrt {16 - x^2 - y^2}$ and z = $\sqrt {4 - x^2 - y^2}$ .

I'm having trouble finding the bounds, would it be

$<br /> \int^{2\pi} _{0} \int^{\frac{\pi}{4}} _{0} \int^{4} _{0}<br /> d\rho d\theta d\phi$

**Scott H** · November 2nd 2009, 12:31 PM

In spherical coordinates, $\rho$ is the distance from the origin, $\theta$ is the angle of the projection on the $xy$ -plane with the $x$ -axis, and $\phi$ is the angle from the $z$ -axis. This is relating points to the Cartesian coordinate system, of course.

Our region is a section of a cone between two spheres. Because this cone, defined by

$z=\sqrt{x^2+y^2},$

makes an angle of $45^{\circ}$ or $\frac{\pi}{4}$ radians with the $z$ -axis, our $d\phi$ limits will be $\left(0,\frac{\pi}{4}\right)$ . Similarly, our $d\theta$ limits will be $(0,2\pi)$ and our $\rho$ limits will be $(2,4)$ .

Now, an important thing to consider is that incrementally-defined volume regions in spherical coordinates grow larger as $\rho$ grows larger, and also grow smaller as $\phi$ decreases to $0$ (as the volume regions shrink to wedges near the $z$ -axis). Nothing like this happens with uniformly distributed Cartesian coordinates. To make up for it, we introduce the factor $\rho^2\sin \phi$ when calculating triple integrals in spherical coordinates:

$\int_0^{\pi/4}\int_0^{2\pi}\int_2^4 \rho^2\sin\phi\,d\rho\,d\theta\,d\phi.$

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