Find the volume of the following solid.
The solid above the region Requals=​{(x,y): 0less than or equals≤xless than or equals≤22​, 0less than or equals≤yless than or equals≤22minus−nothing​x} bounded by the paraboloids zequals=nothingx squaredx2plus+nothingy squaredy2 and zequals=88minus−nothingx squaredx2minus−nothingy squaredy2 and the coordinate planes in the first octant
z equals 8 minus x squared minus y squaredz=8−x2−y2
z equals x squared plus y squaredz=x2+y2

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Part 1
The volume of the region is enter your response here
▼
square units.
units.
cubic units.
​(Type an exact​ answer.)

We are asked to find the volume of the solid bounded by two paraboloids: \( z = x^2 + y^2 \) and \( z = 8 - x^2 - y^2 \), above the region \( R = \{(x, y) \, : \, 0 \leq x \leq \sqrt{2}, \, 0 \leq y \leq \sqrt{2 - x^2} \} \). This is in the first octant of the coordinate system.

### Step 1: Set up the equations of the paraboloids
- The lower paraboloid is given by \( z_1 = x^2 + y^2 \).
- The upper paraboloid is given by \( z_2 = 8 - x^2 - y^2 \).

The volume of the solid is the region between these two surfaces. We need to compute the volume by setting up a double integral over the appropriate region.

### Step 2: Set up the limits of integration

Since the region is symmetric and confined to the first octant, we can use polar coordinates to simplify the integration. The conversion from Cartesian to polar coordinates is given by:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- \( x^2 + y^2 = r^2 \)

The boundaries for \( r \) and \( \theta \) are as follows:
- \( r \) will range from 0 to 2 (as the outer paraboloid reaches 0 when \( r = 2 \)).
- \( \theta \) will range from 0 to \( \pi/2 \), since we're in the first octant.

### Step 3: Set up the double integral

The volume \( V \) is given by the double integral of the difference between the two surfaces:

\[
V = \int_0^{\pi/2} \int_0^2 \left[ (8 - r^2) - r^2 \right] r \, dr \, d\theta
\]

Simplifying the expression inside the integral:

\[
V = \int_0^{\pi/2} \int_0^2 (8 - 2r^2) r \, dr \, d\theta
\]

### Step 4: Solve the integral

First, evaluate the inner integral with respect to \( r \):

\[
\int_0^2 (8r - 2r^3) \, dr
\]

- The integral of \( 8r \) is \( 4r^2 \).
- The integral of \( -2r^3 \) is \( -\frac{r^4}{2} \).

Evaluating these from 0 to 2:

\[
\left[ 4(2)^2 - \frac{(2)^4}{2} \right] - \left[ 4(0)^2 - \frac{(0)^4}{2} \right]
\]
\[
= (4 \times 4 - \frac{16}{2}) = 16 - 8 = 8
\]

Now, evaluate the outer integral with respect to \( \theta \):

\[
\int_0^{\pi/2} 8 \, d\theta = 8 \times \frac{\pi}{2} = 4\pi
\]

### Step 5: Final answer

Thus, the volume of the solid is \( 4\pi \) cubic units.

Would you like more details on any part of this solution? Here are some related questions:

1. How do we convert Cartesian to polar coordinates?
2. What is the geometric interpretation of the volume between two surfaces?
3. How does symmetry simplify integrals in polar coordinates?
4. What are the bounds of integration for regions in the first octant?
5. How does the Jacobian factor \( r \) arise in polar coordinates?

**Tip:** Using polar coordinates simplifies many integrals involving circular symmetry, as it reduces complexity in bounds and expressions.

Find the volume of the following solid. The solid is above the region R = {(x,y): 0 ≤ x ≤ √2, 0 ≤ y ≤ √(2 - x^2)}, bounded by the paraboloids z = x^2 + y^2 and z = 8 - x^2 - y^2, and the coordinate planes in the first octant.

This problem involves calculating the volume of a solid bounded by two paraboloids, z = x^2 + y^2 and z = 8 - x^2 - y^2, in the first octant using polar coordinates and double integrals. A detailed step-by-step solution demonstrates the setup of limits of integration and the evaluation of integrals. Suitable for students studying multivariable calculus.

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