Math Problem Statement

Set up a (sum of) triple integral(s) equal to the mass of the given solid. The solid bounded by z = p 16 − x 2 − y 2, the xy-plane and the xz-plane with density function f(x, y, z) = x 2 y 2 z 2 .

Solution

To set up the integral for the mass of the solid bounded by the surfaces z=16x2y2z = \sqrt{16 - x^2 - y^2}, the xyxy-plane (z=0z = 0), and the xzxz-plane (y=0y = 0), with density function f(x,y,z)=x2y2z2f(x, y, z) = x^2 y^2 z^2, we’ll follow these steps:

1. Analyze the Boundaries

  • The surface z=16x2y2z = \sqrt{16 - x^2 - y^2} represents the upper boundary of the solid, meaning zz ranges from 00 to 16x2y2\sqrt{16 - x^2 - y^2}.
  • The region in the xyxy-plane that this solid projects onto is the circular region defined by x2+y216x^2 + y^2 \leq 16.
  • Since we are bounded by the xzxz-plane (y=0y = 0), the variable yy will range from 00 to 16x2\sqrt{16 - x^2}.

2. Set Up the Triple Integral

The mass MM of the solid is given by the integral: M=solidf(x,y,z)dVM = \iiint\limits_{\text{solid}} f(x, y, z) \, dV where f(x,y,z)=x2y2z2f(x, y, z) = x^2 y^2 z^2.

In Cartesian Coordinates:

We can set up the integral in Cartesian coordinates: M=x=44y=016x2z=016x2y2x2y2z2dzdydxM = \int_{x=-4}^{4} \int_{y=0}^{\sqrt{16 - x^2}} \int_{z=0}^{\sqrt{16 - x^2 - y^2}} x^2 y^2 z^2 \, dz \, dy \, dx where:

  • zz ranges from 00 to 16x2y2\sqrt{16 - x^2 - y^2},
  • yy ranges from 00 to 16x2\sqrt{16 - x^2},
  • xx ranges from 4-4 to 44.

3. Alternative in Cylindrical Coordinates

Alternatively, we can use cylindrical coordinates:

  • Set x=rcosθx = r \cos \theta, y=rsinθy = r \sin \theta, and z=zz = z.
  • The density function becomes f(x,y,z)=(rcosθ)2(rsinθ)2z2=r4cos2θsin2θz2f(x, y, z) = (r \cos \theta)^2 (r \sin \theta)^2 z^2 = r^4 \cos^2 \theta \sin^2 \theta z^2.
  • The region for rr is 0r40 \leq r \leq 4, θ\theta is 0θπ20 \leq \theta \leq \frac{\pi}{2}, and zz is 0z16r20 \leq z \leq \sqrt{16 - r^2}.

Thus, the integral in cylindrical coordinates is: M=θ=0π/2r=04z=016r2r4cos2θsin2θz2rdzdrdθM = \int_{\theta=0}^{\pi/2} \int_{r=0}^{4} \int_{z=0}^{\sqrt{16 - r^2}} r^4 \cos^2 \theta \sin^2 \theta \, z^2 \, r \, dz \, dr \, d\theta where we include an additional rr from the Jacobian determinant.

This setup provides the integral required to find the mass of the solid.

Would you like further details or have any questions on this setup?


Related Questions

  1. How do we evaluate the integral for mass step-by-step?
  2. How can we set up similar integrals for solids bounded by other parabolic surfaces?
  3. What are the benefits of using cylindrical over Cartesian coordinates in such integrals?
  4. How do boundary planes affect the limits of integration in multivariable calculus?
  5. What changes if the density function is dependent only on one variable, say f(x,y,z)=x2f(x, y, z) = x^2?

Tip

In triple integrals, always check the geometry and symmetry of the solid, as these can simplify both the setup and evaluation.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Triple Integrals
Solid Boundaries
Cartesian Coordinates
Cylindrical Coordinates
Density Function

Formulas

Mass of solid: M = ∭_solid f(x, y, z) dV
Density function: f(x, y, z) = x^2 y^2 z^2
Conversion to cylindrical coordinates: x = r cos θ, y = r sin θ, z = z

Theorems

Triple Integral for Mass Calculation
Use of Coordinate Transformations

Suitable Grade Level

Undergraduate Calculus