Questions: Use a triple integral to find the volume of the solid bounded below by the cone z = √(x² + y²) and bounded above by the sphere x² + y² + z² = 338. ∫∫∫ dV = 4π ∫₀^√338 r² (√(338-r²) - r) dr

Use a triple integral to find the volume of the solid bounded below by the cone z = √(x² + y²) and bounded above by the sphere x² + y² + z² = 338.

∫∫∫ dV = 4π ∫₀^√338 r² (√(338-r²) - r) dr
Transcript text: Use a triple integral to find the volume of the solid bounded below by the cone z = √(x² + y²) and bounded above by the sphere x² + y² + z² = 338. \int\int\int dV = 4\pi \int_0^{\sqrt{338}} r^2 (\sqrt{338-r^2} - r) dr
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Solution

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Solution Steps

Step 1: Identify the given surfaces

The problem involves finding the volume of the solid bounded below by the cone z=x2+y2 z = \sqrt{x^2 + y^2} and bounded above by the sphere x2+y2+z2=338 x^2 + y^2 + z^2 = 338 .

Step 2: Convert to cylindrical coordinates

To simplify the integration, convert the given equations to cylindrical coordinates:

  • The cone: z=r z = r
  • The sphere: r2+z2=338 r^2 + z^2 = 338
Step 3: Determine the limits of integration

In cylindrical coordinates:

  • r r ranges from 0 to the intersection of the cone and the sphere.
  • θ \theta ranges from 0 to 2π 2\pi .
  • z z ranges from the cone z=r z = r to the sphere z=338r2 z = \sqrt{338 - r^2} .
Step 4: Set up the triple integral

The volume V V is given by: V=02π0338r338r2rdzdrdθ V = \int_{0}^{2\pi} \int_{0}^{\sqrt{338}} \int_{r}^{\sqrt{338 - r^2}} r \, dz \, dr \, d\theta

Step 5: Integrate with respect to z z

Integrate the innermost integral with respect to z z : r338r2rdz=r[z]r338r2=r(338r2r) \int_{r}^{\sqrt{338 - r^2}} r \, dz = r \left[ z \right]_{r}^{\sqrt{338 - r^2}} = r \left( \sqrt{338 - r^2} - r \right)

Step 6: Integrate with respect to r r

Now, integrate with respect to r r : 0338r(338r2r)dr \int_{0}^{\sqrt{338}} r \left( \sqrt{338 - r^2} - r \right) dr

Step 7: Simplify the integral

Separate the integral into two parts: 0338r338r2dr0338r2dr \int_{0}^{\sqrt{338}} r \sqrt{338 - r^2} \, dr - \int_{0}^{\sqrt{338}} r^2 \, dr

Step 8: Solve the integrals

Use substitution for the first integral and direct integration for the second:

  1. For 0338r338r2dr \int_{0}^{\sqrt{338}} r \sqrt{338 - r^2} \, dr , let u=338r2 u = 338 - r^2 , then du=2rdr du = -2r \, dr .
  2. For 0338r2dr \int_{0}^{\sqrt{338}} r^2 \, dr , use the power rule.
Step 9: Combine the results

Evaluate the integrals and combine the results to find the volume.

Final Answer

The volume of the solid bounded by the cone z=x2+y2 z = \sqrt{x^2 + y^2} and the sphere x2+y2+z2=338 x^2 + y^2 + z^2 = 338 is: V=338338π6 V = \frac{338\sqrt{338}\pi}{6}

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