Question

use spherical coordinates.evaluate ∫∫∫B(x2+y2+z2)2 dv, where b is the ball with center the origin and radius 3.

Answer 1

To solve the integral \(\int\int\int_B (x^2+y^2+z^2)^2 \, dv\) over a sphere with radius 3 using spherical coordinates, you first define the volume element in spherical coordinates and then perform the integration over the respective bounds for r, \(\theta\), and \(\phi\).

Step-by-step explanation:

To evaluate the triple integral \(\int\int\int_B (x^2+y^2+z^2)^2 \, dv\), where B is the ball with center at the origin and radius 3, it is efficient to use spherical coordinates. In spherical coordinates, a point in space is represented by three coordinates: the radius r, the polar angle \(\theta\), and the azimuthal angle \(\phi\). The volume element in spherical coordinates is dv = r^2 \sin(\theta) \, dr \, d\theta \, d\phi.

Given the symmetry of the problem and the nature of the integrand, which is already a square of the radial distance from the origin, the integral simplifies significantly in spherical coordinates:

\(\int\int\int_B (x^2+y^2+z^2)^2 \, dv = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{3} (r^2)^2 \cdot r^2 \sin(\theta) \, dr \, d\theta \, d\phi\)

By evaluating this integral, we are adding up the values of (x^2+y^2+z^2)^2 over the entire volume of a sphere of radius 3. To solve the integral, we first integrate with respect to r from 0 to 3, then \(\theta\) from 0 to \(\pi\), and finally \(\phi\) from 0 to \(2\pi\).

Answer 2

The integral ∫∫∫B (x2+y2+z2)2 dv over a sphere with radius 3 is simplified using spherical coordinates and performing the triple integral step by step, integrating over the coordinates φ, θ, and r.

Step-by-step explanation:

To evaluate the triple integral ∫∫∫B (x2+y2+z2)2 dv, where B is the ball with center the origin and radius 3 using spherical coordinates, we need to express the volume element dv and the function in spherical coordinates. In spherical coordinates, x2+y2+z2 is simply the radius squared r2, and the volume element dv is r2sin(θ)drdθdφ.

The bounds for the integral in spherical coordinates will be from 0 to 3 for r, 0 to π for θ, and 0 to 2π for φ. The triple integral therefore simplifies to: ∫03∫0π∫02π r4sin(θ)drdθdφ.

Performing the integration step by step, we first integrate with respect to φ, which is straightforward since it does not have any variable dependent on φ. Next, we integrate with respect to θ, which involves integrating sin(θ), and finally, we integrate with respect to r being cognizant that this is a polynomial in r.