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To solve the problem of finding the volume of the solid resulting from rotating the curve

\[
y = \frac{9}{x^2 + 3x + 2}
\]

from \(x = 0\) to \(x = 1\), we need to address both rotations: (a) around the x-axis and (b) around the y-axis.

### (a) Volume of revolution around the x-axis:

To calculate the volume when rotating around the x-axis, we use the disk method. The formula is:

\[
V = \pi \int_{a}^{b} \left( y(x) \right)^2 dx
\]

Substitute \( y(x) = \frac{9}{x^2 + 3x + 2} \), and the limits are from \( x = 0 \) to \( x = 1 \).

\[
V = \pi \int_{0}^{1} \left( \frac{9}{x^2 + 3x + 2} \right)^2 dx
\]

Now, I will compute this integral.The volume of the solid obtained by rotating the curve around the x-axis is:

\[
V = 54\pi (\ln\left(\frac{27}{64}\right) + 1)
\]

### (b) Volume of revolution around the y-axis:

To find the volume when rotating around the y-axis, we use the shell method. The formula is:

\[
V = 2\pi \int_{a}^{b} x \cdot y(x) \, dx
\]

Substitute \( y(x) = \frac{9}{x^2 + 3x + 2} \), and integrate from \(x = 0\) to \(x = 1\):

\[
V = 2\pi \int_{0}^{1} x \cdot \frac{9}{x^2 + 3x + 2} \, dx
\]

Let me compute this integral next.The volume of the solid obtained by rotating the curve around the y

Find the volume of the resulting solid if the region under the curve y = 9 / (x^2 + 3x + 2) from x = 0 to x = 1 is rotated about the x-axis and the y-axis.

Learn how to find the volume of the solid resulting from rotating the region under the curve y = 9 / (x^2 + 3x + 2) from x = 0 to x = 1 around the x-axis and y-axis. This problem involves calculus concepts like the Disk Method and Shell Method, suitable for college-level students.

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