Math Problem Statement

Express the volume of the solid enclosed by the paraboloids y=2x2+2z2 and y=80−3x2−3z2 with a triple integral.

Given the region of integration which would be the easiest order of integration? Check all that apply.

Solution

To express the volume of the solid enclosed by the paraboloids y=2x2+2z2y = 2x^2 + 2z^2 and y=803x23z2y = 80 - 3x^2 - 3z^2, we first need to set up the triple integral for this region.

Step 1: Find the Bounds

  1. Intersection of Paraboloids: Set y=2x2+2z2y = 2x^2 + 2z^2 equal to y=803x23z2y = 80 - 3x^2 - 3z^2: 2x2+2z2=803x23z22x^2 + 2z^2 = 80 - 3x^2 - 3z^2 Simplifying this equation: 5x2+5z2=80x2+z2=165x^2 + 5z^2 = 80 \Rightarrow x^2 + z^2 = 16 This implies that the projection of the region onto the xzxz-plane is a circle of radius 4 (since x2+z216x^2 + z^2 \leq 16).

  2. Volume Integral: The volume of the region can be represented by the triple integral V=region1dVV = \iiint\limits_{\text{region}} 1 \, dV where the region is bounded by y=2x2+2z2y = 2x^2 + 2z^2 from below and y=803x23z2y = 80 - 3x^2 - 3z^2 from above.

Step 2: Set up the Integral

We can set up the triple integral as follows:

V=x=44z=16x216x2y=2x2+2z2803x23z21dydzdxV = \int_{x=-4}^{4} \int_{z=-\sqrt{16 - x^2}}^{\sqrt{16 - x^2}} \int_{y = 2x^2 + 2z^2}^{80 - 3x^2 - 3z^2} 1 \, dy \, dz \, dx

Choosing the Easiest Order of Integration

Given the symmetry in the xx- and zz-variables (as indicated by x2+z216x^2 + z^2 \leq 16), it would be beneficial to switch to cylindrical coordinates for easier integration.

In cylindrical coordinates:

  • Let x=rcosθx = r \cos \theta and z=rsinθz = r \sin \theta,
  • The bounds become r[0,4]r \in [0, 4], θ[0,2π]\theta \in [0, 2\pi],
  • yy-bounds remain as y=2r2y = 2r^2 and y=803r2y = 80 - 3r^2.

Thus, the integral becomes: V=θ=02πr=04y=2r2803r2rdydrdθV = \int_{\theta=0}^{2\pi} \int_{r=0}^{4} \int_{y = 2r^2}^{80 - 3r^2} r \, dy \, dr \, d\theta

Easiest Order of Integration

The cylindrical coordinate system simplifies the problem, making the integration order dθdrdyd\theta \, dr \, dy the easiest. Thus, check all that apply for this problem would include:

  1. dθdrdyd\theta \, dr \, dy – preferred in cylindrical coordinates due to radial symmetry.

Would you like further details on each step or have any specific questions?


  1. How do you derive the bounds for rr and θ\theta in cylindrical coordinates?
  2. What is the reason for choosing cylindrical coordinates here?
  3. Can the integral be solved in Cartesian coordinates, and how complex would it be?
  4. What are the limits for yy in this integral, and how do they relate to the paraboloids?
  5. How does the radial symmetry of the problem simplify the integration?

Tip: When dealing with circular or radial symmetry in 3D regions, cylindrical or spherical coordinates often simplify the setup and integration.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Triple Integrals
Cylindrical Coordinates
Volume of Solids

Formulas

Triple integral for volume: V = ∫∫∫_Region 1 dV
Cylindrical coordinates: x = r cos(θ), z = r sin(θ), y = y

Theorems

Fubini's Theorem
Volume by Integration

Suitable Grade Level

College Level