Math Problem Statement

Use spherical coordinates. Evaluate

E (x2 + y2) dV, where E lies between the spheres x2 + y2 + z2 = 1 and x2 + y2 + z2 = 16.

Solution

To evaluate the integral

E(x2+y2)dV\iiint_E (x^2 + y^2) \, dV

where EE lies between the spheres x2+y2+z2=1x^2 + y^2 + z^2 = 1 and x2+y2+z2=16x^2 + y^2 + z^2 = 16, we can use spherical coordinates.

Step 1: Setting up the spherical coordinates

In spherical coordinates:

  • x=ρsinθcosϕx = \rho \sin \theta \cos \phi
  • y=ρsinθsinϕy = \rho \sin \theta \sin \phi
  • z=ρcosθz = \rho \cos \theta
  • The volume element dVdV in spherical coordinates is given by dV=ρ2sinθdρdθdϕdV = \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi.

In these coordinates, x2+y2=ρ2sin2θx^2 + y^2 = \rho^2 \sin^2 \theta and the equation for the spheres becomes:

  • Inner sphere: ρ=1\rho = 1
  • Outer sphere: ρ=4\rho = 4

Thus, ρ\rho ranges from 1 to 4, θ\theta from 0 to π\pi, and ϕ\phi from 0 to 2π2\pi.

Step 2: Set up the integral

The integral now becomes

E(x2+y2)dV=02π0π14ρ2sin2θρ2sinθdρdθdϕ\iiint_E (x^2 + y^2) \, dV = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{1}^{4} \rho^2 \sin^2 \theta \cdot \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi

Simplify the integrand:

=02π0π14ρ4sin3θdρdθdϕ= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{1}^{4} \rho^4 \sin^3 \theta \, d\rho \, d\theta \, d\phi

Step 3: Separate the integrals

Since the integrand is separable in terms of ρ\rho, θ\theta, and ϕ\phi, we can write:

02πdϕ0πsin3θdθ14ρ4dρ\int_{0}^{2\pi} \, d\phi \int_{0}^{\pi} \sin^3 \theta \, d\theta \int_{1}^{4} \rho^4 \, d\rho

  1. Evaluate the ϕ\phi-integral:

    02πdϕ=2π\int_{0}^{2\pi} \, d\phi = 2\pi

  2. Evaluate the ρ\rho-integral:

    14ρ4dρ=[ρ55]14=455155=1024515=10235\int_{1}^{4} \rho^4 \, d\rho = \left[ \frac{\rho^5}{5} \right]_{1}^{4} = \frac{4^5}{5} - \frac{1^5}{5} = \frac{1024}{5} - \frac{1}{5} = \frac{1023}{5}

  3. Evaluate the θ\theta-integral:

    0πsin3θdθ\int_{0}^{\pi} \sin^3 \theta \, d\theta

To solve this, use the identity sin3θ=sinθ(1cos2θ)\sin^3 \theta = \sin \theta (1 - \cos^2 \theta) and then apply integration techniques, or use a table of integrals. The result is:

0πsin3θdθ=43\int_{0}^{\pi} \sin^3 \theta \, d\theta = \frac{4}{3}

Step 4: Combine the results

Putting it all together:

E(x2+y2)dV=2π4310235=8196π15\iiint_E (x^2 + y^2) \, dV = 2\pi \cdot \frac{4}{3} \cdot \frac{1023}{5} = \frac{8196\pi}{15}

Thus, the evaluated integral is:

8196π15\boxed{\frac{8196\pi}{15}}

Would you like more details or have questions about any specific part of the solution?


Here are some related questions:

  1. How are spherical coordinates derived from Cartesian coordinates?
  2. What applications require the use of spherical coordinates?
  3. How would the limits change if the region were bounded by a cylinder instead of spheres?
  4. Why does the volume element dV=ρ2sinθdρdθdϕdV = \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi include ρ2\rho^2 and sinθ\sin \theta?
  5. What other coordinate systems are commonly used in multivariable calculus, and how do they compare to spherical coordinates?

Tip: In spherical coordinates, always carefully analyze the bounds of ρ\rho, θ\theta, and ϕ\phi, as they vary based on the region of integration.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Spherical Coordinates
Volume Integrals

Formulas

Volume element in spherical coordinates: \( dV = \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi \)
Coordinate conversions: \( x = \rho \sin \theta \cos \phi \), \( y = \rho \sin \theta \sin \phi \), \( z = \rho \cos \theta \)

Theorems

Integration in Spherical Coordinates
Fubini's Theorem for Triple Integrals

Suitable Grade Level

College (Calculus III or Multivariable Calculus)