Math Problem Statement

The solid enclosed by the paraboloid x = y^2 + z^2 and the plane x = 16

Solution

To find the volume of the solid enclosed by the paraboloid x=y2+z2x = y^2 + z^2 and the plane x=16x = 16, we can set up an integral in cylindrical coordinates. In cylindrical coordinates (r,θ,x)(r, \theta, x), we have:

  1. The paraboloid x=y2+z2x = y^2 + z^2 can be rewritten as x=r2x = r^2.
  2. The plane x=16x = 16 gives us an upper bound for xx.

Steps:

  1. Set up the bounds:

    • For xx, we have r2x16r^2 \leq x \leq 16.
    • For rr, since x=r216x = r^2 \leq 16, we find r16=4r \leq \sqrt{16} = 4. So 0r40 \leq r \leq 4.
    • For θ\theta, it goes around the entire circle, so 0θ2π0 \leq \theta \leq 2\pi.
  2. Volume Integral: The volume VV of the solid is given by: V=02π04r216rdxdrdθV = \int_{0}^{2\pi} \int_{0}^{4} \int_{r^2}^{16} r \, dx \, dr \, d\theta Here, the rr factor appears because of the cylindrical coordinate volume element dV=rdxdrdθdV = r \, dx \, dr \, d\theta.

  3. Evaluate the Integral:

    • First, integrate with respect to xx: r216rdx=r(16r2)\int_{r^2}^{16} r \, dx = r \cdot (16 - r^2)

    • Substitute back into the integral: V=02π04r(16r2)drdθV = \int_{0}^{2\pi} \int_{0}^{4} r(16 - r^2) \, dr \, d\theta

    • Separate the integrals: V=02πdθ04(16rr3)drV = \int_{0}^{2\pi} d\theta \int_{0}^{4} (16r - r^3) \, dr

    • Integrate with respect to rr: 04(16rr3)dr=[8r2r44]04=(8162564)=12864=64\int_{0}^{4} (16r - r^3) \, dr = \left[ 8r^2 - \frac{r^4}{4} \right]_{0}^{4} = \left( 8 \cdot 16 - \frac{256}{4} \right) = 128 - 64 = 64

    • Now, integrate with respect to θ\theta: V=02π64dθ=642π=128πV = \int_{0}^{2\pi} 64 \, d\theta = 64 \cdot 2\pi = 128\pi

Final Answer:

The volume of the solid enclosed by the paraboloid and the plane is: V=128πV = 128\pi

Would you like more details or have any questions on specific steps?


Here are some related questions:

  1. How would the solution change if the paraboloid were rotated along a different axis?
  2. What would the volume be if the plane were at x=25x = 25 instead?
  3. Can we solve this problem using Cartesian coordinates instead of cylindrical?
  4. How would the problem change if the paraboloid had the equation x=y2+z2+1x = y^2 + z^2 + 1?
  5. What is the general formula for the volume of solids bounded by paraboloids and planes?

Tip: Always check whether switching to a different coordinate system (such as cylindrical) simplifies setting up integrals for symmetric regions.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Volume of Solids
Cylindrical Coordinates

Formulas

Volume integral formula V = ∫∫∫ dV
Cylindrical coordinate transformation: x = r^2

Theorems

Fubini's Theorem for multiple integrals

Suitable Grade Level

College-level Calculus