Math Problem Statement

Q2. Find the volume of the solid bounded by x=8-y^2-z^2 and x=y^2+z^2.

Solution

To find the volume of the solid bounded by the surfaces x=8y2z2x = 8 - y^2 - z^2 and x=y2+z2x = y^2 + z^2, we will use cylindrical coordinates.

First, let's express the given equations in cylindrical coordinates:

x=8r2x = 8 - r^2

x=r2x = r^2

where r2=y2+z2r^2 = y^2 + z^2.

In cylindrical coordinates, the volume integral is:

V=dV=rdzdrdθV = \int \int \int dV = \int \int \int r \, dz \, dr \, d\theta

The surfaces intersect where 8r2=r28 - r^2 = r^2, so:

8=2r2    r2=4    r=28 = 2r^2 \implies r^2 = 4 \implies r = 2

The bounds for rr are from 0 to 2, for θ\theta from 0 to 2π2\pi, and for zz from the lower surface z=8xz = -\sqrt{8 - x} to the upper surface z=8xz = \sqrt{8 - x}.

We need to integrate x=r2x = r^2 to x=8r2x = 8 - r^2:

V=02π02r28r2rdxdrdθV = \int_0^{2\pi} \int_0^2 \int_{r^2}^{8-r^2} r \, dx \, dr \, d\theta

However, since we are using cylindrical coordinates and xx is replaced by the equation for the height differential dzdz:

V=02π02r28r2rdzdrdθV = \int_0^{2\pi} \int_0^2 \int_{r^2}^{8-r^2} r \, dz \, dr \, d\theta

We need to be cautious here; dzdz range is 28r22 \sqrt{8 - r^2}:

V=02π02(82r2)rdrdθV = \int_0^{2\pi} \int_0^2 (8 - 2r^2) \, r \, dr \, d\theta

Solving the integral:

  1. Integrate with respect to rr:

V=02π02(8r2r3)drdθV = \int_0^{2\pi} \int_0^2 (8r - 2r^3) \, dr \, d\theta

=02π[4r2r42]02dθ= \int_0^{2\pi} \left[ 4r^2 - \frac{r^4}{2} \right]_0^2 \, d\theta

=02π[4(22)(24)2]dθ= \int_0^{2\pi} \left[ 4(2^2) - \frac{(2^4)}{2} \right] \, d\theta

=02π[168]dθ= \int_0^{2\pi} \left[ 16 - 8 \right] \, d\theta

=02π8dθ= \int_0^{2\pi} 8 \, d\theta

  1. Integrate with respect to θ\theta:

V=8θ02π=8×2π=16πV = 8 \theta \bigg|_0^{2\pi} = 8 \times 2\pi = 16\pi

Therefore, the volume of the solid is 16π\boxed{16\pi}.

Would you like further details or have any questions?

Here are 8 related questions for further exploration:

  1. How would the volume change if the bounds for xx were modified?
  2. What is the geometric interpretation of the solid formed by these bounds?
  3. How can this problem be solved using Cartesian coordinates?
  4. What are some applications of cylindrical coordinates in physics?
  5. How would the volume change if the functions were shifted?
  6. How do you convert integrals from Cartesian to cylindrical coordinates?
  7. What other shapes could be formed using similar bounded surfaces?
  8. How does symmetry help in solving these types of integrals?

Tip: When converting integrals from Cartesian to cylindrical coordinates, always make sure to include the Jacobian determinant rr to account for the area element in polar coordinates.

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Math Problem Analysis

Mathematical Concepts

Volume of solids
Cylindrical coordinates
Triple integrals

Formulas

Volume integral in cylindrical coordinates

Theorems

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Suitable Grade Level

College Level