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
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
Solution
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 and bounded above by the sphere x2+y2+z2=338.
Step 2: Convert to cylindrical coordinates
To simplify the integration, convert the given equations to cylindrical coordinates:
The cone: z=r
The sphere: r2+z2=338
Step 3: Determine the limits of integration
In cylindrical coordinates:
r ranges from 0 to the intersection of the cone and the sphere.
θ ranges from 0 to 2π.
z ranges from the cone z=r to the sphere z=338−r2.
Step 4: Set up the triple integral
The volume V is given by:
V=∫02π∫0338∫r338−r2rdzdrdθ
Step 5: Integrate with respect to z
Integrate the innermost integral with respect to z:
∫r338−r2rdz=r[z]r338−r2=r(338−r2−r)
Step 6: Integrate with respect to r
Now, integrate with respect to r:
∫0338r(338−r2−r)dr
Step 7: Simplify the integral
Separate the integral into two parts:
∫0338r338−r2dr−∫0338r2dr
Step 8: Solve the integrals
Use substitution for the first integral and direct integration for the second:
For ∫0338r338−r2dr, let u=338−r2, then du=−2rdr.
For ∫0338r2dr, 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 and the sphere x2+y2+z2=338 is:
V=6338338π