Problem. Let R be the region bounded above by z = 8 - x2 – y? and below by z = x² + y?. Then the volume of R is:

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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**Problem 1:**

Let \( R \) be the region bounded above by \( z = 8 - x^2 - y^2 \) and below by \( z = x^2 + y^2 \). Then the volume of \( R \) is: \(\boxed{?} \)

**Problem 2:**

Let \( R \) be the portion of \( x^2 + y^2 + z^2 = 4 \) above the plane \( z=1 \). Consider the integral \(\iiint_R z \, dV \).

- **In cylindrical coordinates, the integral would be:**

  \[
  \int_{\boxed{?}}^{\boxed{?}} \int_{\boxed{?}}^{\boxed{?}} \int_{\boxed{?}}^{\boxed{?}} z \, dz \, dr \, d\theta.
  \]

- **In spherical coordinates, the integral would be (input "p" for phi):**

  \[
  \int_{\boxed{?}}^{\boxed{?}} \int_{\boxed{?}}^{\boxed{?}} \int_{\boxed{?}}^{\boxed{?}} \rho^3 \cos(p) \sin(p) \, d\rho \, dp \, d\theta.
  \]

- **Evaluating either way gives an answer of:** \(\boxed{?} \)
Transcribed Image Text:**Problem 1:** Let \( R \) be the region bounded above by \( z = 8 - x^2 - y^2 \) and below by \( z = x^2 + y^2 \). Then the volume of \( R \) is: \(\boxed{?} \) **Problem 2:** Let \( R \) be the portion of \( x^2 + y^2 + z^2 = 4 \) above the plane \( z=1 \). Consider the integral \(\iiint_R z \, dV \). - **In cylindrical coordinates, the integral would be:** \[ \int_{\boxed{?}}^{\boxed{?}} \int_{\boxed{?}}^{\boxed{?}} \int_{\boxed{?}}^{\boxed{?}} z \, dz \, dr \, d\theta. \] - **In spherical coordinates, the integral would be (input "p" for phi):** \[ \int_{\boxed{?}}^{\boxed{?}} \int_{\boxed{?}}^{\boxed{?}} \int_{\boxed{?}}^{\boxed{?}} \rho^3 \cos(p) \sin(p) \, d\rho \, dp \, d\theta. \] - **Evaluating either way gives an answer of:** \(\boxed{?} \)
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