# Triple Integral in Rectangular Coordinates Converting to Spherical Coordinates

• donald17
In summary, the homework statement is that given a spherical region, the boundaries for the theta and phi limits can be found using the lower half plane in the xy- plane. The limits for rho and y should also be introduced. The integral for theta is just the volume of the region and can be found using the new coordinates x', y', z'.

## Homework Statement

Given that:

Write an equivalent integral in spherical coordinates.

## Homework Equations

(Triple integral in spherical coordinates.)

(Conversions from rectangular to spherical coordinates.)(What spherical coordinates entail)

## The Attempt at a Solution

The region is going to be the region of the lower part of a sphere with radius a moved up a units on the z-axis that lies in the -x and -y plane. The limits of integration for theta should be 0 to pi/2. Need assistance with limits for phi and rho.

Last edited by a moderator:
y ranges from -a to 0 so we are talking about the lower half plane in the xy- plane.

For each y, x ranges from $-\sqrt{a^2- y^2}$ to 0. $x= -\sqrt{a^2- y^2}$ gives $x^2+ y^2= a^2$, the circle with center (0,0) and radius a. Since x is the negative square root and we already know that we are in the lower half plane (y< 0), we are looking at the portion of that circle in the third quadrant.

For each (x,y), z ranges from $a- \sqrt{a^2- x^2- y^2}$ to a. $z= a- \sqrt{a^2- x^2- y^2}$ gives $x^2+ y^2+ (z-a)^2= a^2$, the sphere with center at (0,0,a) and radius a. Since z goes up to a, we are in the lower half sphere. That is, we are looking at the part of the sphere $x^2+ y^2+ (z-a)^2= a^2$ where x and y are negative and z< a.

I would be inclined to introduce new coordinates: x'= x, y'= y, z'= z- a. In those coordinates, the region we are integrating over is the "seventh octant" in which x, y, and z are all negative. To get that sphere, let $\rho$ go from 0 to a, $\theta$ from $\pi$ to $3\pi/2$, and $\phi$ from $\pi/2$ to $\pi$. The integrand is still "1" and the "differential of volume" is still $\rho sin^2(\phi)d\phi d\theta d\rho$. In fact, because the integrand is "1", this integral is just the volume of the region: 1/8 of the volume of a sphere of radius a and so is $(1/8)(4/3)\pi a^3= (1/6)\pi a^3$. You can use that as a check.

Shouldn't we be integrating the region of the sphere in the -x, -y, +z plane, since z goes from the bottom of the sphere, moved up a on the z-axis, to the plane z=a?

## 1. What is the purpose of converting a triple integral from rectangular to spherical coordinates?

The purpose of converting a triple integral from rectangular to spherical coordinates is to make the integral easier to evaluate. Spherical coordinates are often better suited for problems with spherical symmetry, such as those involving spheres, cones, or cylinders.

## 2. How do you convert a triple integral from rectangular to spherical coordinates?

To convert a triple integral from rectangular to spherical coordinates, you use the following equations:

x = ρsinφcosθ

y = ρsinφsinθ

z = ρcosφ

where ρ is the distance from the origin to the point, φ is the angle between the positive z-axis and the line connecting the origin to the point, and θ is the angle between the positive x-axis and the projection of the line onto the xy-plane.

## 3. What are the limits of integration when converting to spherical coordinates?

The limits of integration when converting to spherical coordinates are determined by the shape of the region being integrated. They can be found by setting the equations for x, y, and z equal to the limits of integration in the rectangular coordinates.

## 4. How does converting to spherical coordinates affect the integral?

Converting to spherical coordinates can simplify the integral and make it easier to evaluate. This is because in spherical coordinates, the integral is often easier to split into separate parts, each of which can be evaluated separately.

## 5. Are there any disadvantages to converting to spherical coordinates for a triple integral?

One potential disadvantage of converting to spherical coordinates is that the limits of integration may become more complicated. This can make the integral more difficult to evaluate, especially if the region being integrated has a complex shape.