Set up triple integrals for the volume of the sphere

rhoρequals=55

in

Bold a.**a.**

​spherical,

Bold b.**b.**

​cylindrical, and

Bold c. rectangular**c.** rectangular

We are asked to set up triple integrals for the volume of a sphere with radius \( \rho = 5 \) in three different coordinate systems: **spherical**, **cylindrical**, and **rectangular** coordinates.

### a. Spherical Coordinates:
In spherical coordinates, the volume element is given by \( dV = \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi \), where:
- \( \rho \) is the radial distance (from the origin),
- \( \theta \) is the azimuthal angle (in the xy-plane, 0 to \( 2\pi \)),
- \( \phi \) is the polar angle (from the positive z-axis, 0 to \( \pi \)).

The limits for a sphere of radius 5 are:
- \( \rho \) goes from 0 to 5,
- \( \theta \) goes from 0 to \( 2\pi \),
- \( \phi \) goes from 0 to \( \pi \).

The integral for the volume is:

\[
V = \int_0^{2\pi} \int_0^{\pi} \int_0^5 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta
\]

### b. Cylindrical Coordinates:
In cylindrical coordinates, the volume element is given by \( dV = r \, dz \, dr \, d\theta \), where:
- \( r \) is the radial distance in the xy-plane,
- \( \theta \) is the azimuthal angle (0 to \( 2\pi \)),
- \( z \) is the height (from \( -\sqrt{25 - r^2} \) to \( \sqrt{25 - r^2} \), derived from the equation of the sphere \( x^2 + y^2 + z^2 = 25 \)).

The limits are:
- \( r \) goes from 0 to 5,
- \( \theta \) goes from 0 to \( 2\pi \),
- \( z \) goes from \( -\sqrt{25 - r^2} \) to \( \sqrt{25 - r^2} \).

The integral for the volume is:

\[
V = \int_0^{2\pi} \int_0^5 \int_{-\sqrt{25 - r^2}}^{\sqrt{25 - r^2}} r \, dz \, dr \, d\theta
\]

### c. Rectangular Coordinates:
In rectangular coordinates, the volume element is \( dV = dx \, dy \, dz \). The equation of the sphere is \( x^2 + y^2 + z^2 = 25 \), so the limits on \( x \), \( y \), and \( z \) are determined by this equation.

The limits are:
- \( x \) goes from \( -5 \) to \( 5 \),
- \( y \) goes from \( -\sqrt{25 - x^2} \) to \( \sqrt{25 - x^2} \),
- \( z \) goes from \( -\sqrt{25 - x^2 - y^2} \) to \( \sqrt{25 - x^2 - y^2} \).

The integral for the volume is:

\[
V = \int_{-5}^{5} \int_{-\sqrt{25 - x^2}}^{\sqrt{25 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{\sqrt{25 - x^2 - y^2}} dz \, dy \, dx
\]

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Would you like more details or have any questions?

Here are 5 related questions:
1. How do you evaluate the spherical integral for a general radius \( \rho \)?
2. How does the Jacobian affect the volume integral in different coordinates?
3. Can you derive the limits for the cylindrical case from the equation of the sphere?
4. What are the advantages of using spherical coordinates for spherical symmetry?
5. How would the integrals change if the sphere were shifted from the origin?

**Tip:** When dealing with spherical symmetry, spherical coordinates simplify the problem and reduce the complexity of the integrals.

Set up triple integrals for the volume of the sphere rho equals 5 in spherical, cylindrical, and rectangular coordinates.

Learn how to set up triple integrals for the volume of a sphere with radius 5 using spherical, cylindrical, and rectangular coordinates. This problem covers key concepts in multivariable calculus and the transformation of integrals between coordinate systems, suitable for students at the undergraduate level.

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