Find the volume of the solid bounded by the surface 

zequals=​f(x,y) 

and the​ xy-plane.

z equals StartFraction 20 Over 1 plus x squared plus y squared EndFraction minus 2z=201+x2+y2−2


To find the volume of the solid bounded by the surface \( z = \frac{20}{1 + x^2 + y^2} - 2 \) and the \( xy \)-plane, we need to integrate the function over the region in the \( xy \)-plane where \( z \geq 0 \).

### Step 1: Set up the equation
The function given is:
\[
z = \frac{20}{1 + x^2 + y^2} - 2
\]

The solid lies above the \( xy \)-plane, so \( z \geq 0 \), which means:
\[
\frac{20}{1 + x^2 + y^2} - 2 \geq 0
\]
Solving for \( x^2 + y^2 \), we get:
\[
\frac{20}{1 + x^2 + y^2} \geq 2
\]
\[
1 + x^2 + y^2 \leq 10
\]
\[
x^2 + y^2 \leq 9
\]

Thus, the region in the \( xy \)-plane is a disk of radius 3 centered at the origin.

### Step 2: Convert to polar coordinates
Since the region is a disk, it is convenient to convert the problem to polar coordinates. In polar coordinates:
\[
x = r\cos(\theta), \quad y = r\sin(\theta)
\]
The function becomes:
\[
z = \frac{20}{1 + r^2} - 2
\]

The area element \( dA \) in polar coordinates is \( r \, dr \, d\theta \). The volume is given by the double integral:
\[
V = \int_0^{2\pi} \int_0^3 \left( \frac{20}{1 + r^2} - 2 \right) r \, dr \, d\theta
\]

### Step 3: Compute the integral
We first compute the radial part of the integral:
\[
\int_0^3 \left( \frac{20}{1 + r^2} - 2 \right) r \, dr
\]
This can be split into two integrals:
\[
\int_0^3 \frac{20r}{1 + r^2} \, dr - \int_0^3 2r \, dr
\]

The first integral is a standard one, which results in:
\[
\int \frac{r \, dr}{1 + r^2} = \frac{1}{2} \ln(1 + r^2)
\]
So the first integral becomes:
\[
20 \left[ \frac{1}{2} \ln(1 + r^2) \right]_0^3 = 10 \ln(10)
\]

The second integral is straightforward:
\[
\int_0^3 2r \, dr = \left[ r^2 \right]_0^3 = 9
\]

Thus, the radial integral becomes:
\[
10 \ln(10) - 9
\]

### Step 4: Compute the full volume
Now, integrate with respect to \( \theta \):
\[
V = \int_0^{2\pi} \left( 10 \ln(10) - 9 \right) d\theta = 2\pi (10 \ln(10) - 9)
\]

So, the volume of the solid is:
\[
V = 2\pi (10 \ln(10) - 9)
\]

### Final Answer:
The volume of the solid is \( 2\pi (10 \ln(10) - 9) \).

---

Would you like a detailed explanation on any part of the process? Here are some related questions to further explore this topic:

1. How do polar coordinates simplify the calculation of volumes?
2. What is the geometric meaning of integrating over a disk?
3. Can this method be applied to other surfaces bounded by the xy-plane?
4. How would the volume change if the surface equation were different?
5. How does the integral of a rational function like \( \frac{r}{1 + r^2} \) work?

**Tip:** Converting to polar coordinates is often useful when the region of integration is circular or radial in nature.

Find the volume of the solid bounded by the surface z = 20 / (1 + x^2 + y^2) - 2 and the xy-plane.

Learn how to find the volume of a solid bounded by the surface z = 20 / (1 + x^2 + y^2) - 2 and the xy-plane. This calculus problem uses double integration in polar coordinates, covering concepts in multivariable calculus and volume computation for solids.

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