Use the shell method to write and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the *x*-axis. Bound by 0 to 36

*x* + *y*2 = 36




To solve this problem, we need to use the **shell method** to find the volume of the solid generated by revolving the region bounded by the curve \( x + y^2 = 36 \) around the x-axis.

### Step 1: Rearrange the equation in terms of \( y \)

The given equation is:
\[
x + y^2 = 36
\]
Solving for \( x \), we get:
\[
x = 36 - y^2
\]

This is the equation we will use for the shell method.

### Step 2: Shell method setup

The shell method formula for the volume of a solid obtained by rotating a region around the **x-axis** is:

\[
V = 2\pi \int_{y_1}^{y_2} y \cdot f(y) \, dy
\]
Where:
- \( y \) is the radius of the shell (the distance from the x-axis),
- \( f(y) \) is the height of the shell, given by the function \( x = f(y) \),
- \( y_1 \) and \( y_2 \) are the limits of integration (in this case, 0 and 6, as determined by the limits on \( y \)).

The curve intersects the x-axis when \( y = 0 \), and the maximum value of \( y \) occurs when \( x = 0 \), which gives \( y = \sqrt{36} = 6 \).

Thus, the limits of integration are \( y = 0 \) to \( y = 6 \).

### Step 3: Volume integral

Now, applying the shell method:

\[
V = 2\pi \int_0^6 y \cdot (36 - y^2) \, dy
\]

### Step 4: Expand and evaluate the integral

First, expand the integrand:

\[
V = 2\pi \int_0^6 (36y - y^3) \, dy
\]

Now, evaluate each term separately:

\[
V = 2\pi \left[ \int_0^6 36y \, dy - \int_0^6 y^3 \, dy \right]
\]

#### First integral:

\[
\int_0^6 36y \, dy = 36 \cdot \frac{y^2}{2} \Big|_0^6 = 36 \cdot \frac{36}{2} = 648
\]

#### Second integral:

\[
\int_0^6 y^3 \, dy = \frac{y^4}{4} \Big|_0^6 = \frac{6^4}{4} = \frac{1296}{4} = 324
\]

### Step 5: Calculate the final volume

Now substitute the values back:

\[
V = 2\pi (648 - 324) = 2\pi (324) = 648\pi
\]

Thus, the volume of the solid generated by revolving the region around the x-axis is:

\[
V = 648\pi \, \text{cubic units}
\]

### Summary:
The integral that gives the volume is:
\[
V = 2\pi \int_0^6 y(36 - y^2) \, dy
\]
The evaluated volume is:
\[
V = 648\pi \, \text{cubic units}
\]

Would you like more details on any part of the solution?

### Related questions:
1. How would you modify the shell method if the region was rotated about the y-axis?
2. What is the geometrical interpretation of the shell method in this problem?
3. How does the volume change if the region is revolved around the line \(y = -2\)?
4. Could we use the disk method instead of the shell method for this problem? How?
5. What if the bounds were changed to \(x = 0\) to \(x = 36\)? How would the setup change?

#### Tip:
Always sketch the region and axis of rotation to better visualize the geometry of the problem before setting up the integral.

Use the shell method to write and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis. Bound by 0 to 36, x + y^2 = 36.

This page explains how to find the volume of a solid generated by revolving the region bounded by the curve x + y^2 = 36 about the x-axis using the shell method. Detailed steps include setting up and evaluating the integral to arrive at a solution of 648π cubic units.

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