$$ \left. \begin{array} { l }{ V _ { 2 } = \pi \int _ { 9 / 2 } ^ { 5 } ( f ( x ) ^ { 2 } - g ( x ) ^ { 2 } ) d y } \ { = \pi \int _ { 9 / 2 } ^ { 5 } ( ( 1 / 2 ) ^ { 2 } - ( x - 9 / 2 ) ^ { 2 } ) d x } \end{array} ight. $$

Question

$$ \left. \begin{array} { l }{ V _ { 2 } = \pi \int _ { 9 / 2 } ^ { 5 } ( f ( x ) ^ { 2 } - g ( x ) ^ { 2 } ) d y } \ { = \pi \int _ { 9 / 2 } ^ { 5 } ( ( 1 / 2 ) ^ { 2 } - ( x - 9 / 2 ) ^ { 2 } ) d x } \end{array} ight. $$

UpStudy AI · Accepted Answer

Certainly! Let's evaluate the given integral step by step to find the volume $$ V_2 $$.

### Given Expression

$$
V_2 = \pi \int_{9/2}^{5} \left( f(x)^2 - g(x)^2 \right) \, dx
$$

Where:
- $$ f(x) = \frac{1}{2} $$
- $$ g(x) = x - \frac{9}{2} $$

### Substituting the Functions

Plugging $$ f(x) $$ and $$ g(x) $$ into the integral:

$$
V_2 = \pi \int_{9/2}^{5} \left( \left(\frac{1}{2}\right)^2 - \left(x - \frac{9}{2}\right)^2 \right) \, dx
$$

Simplify the expressions inside the integral:

$$
V_2 = \pi \int_{9/2}^{5} \left( \frac{1}{4} - \left(x - \frac{9}{2}\right)^2 \right) \, dx
$$

### Change of Variable

Let’s make a substitution to simplify the integral:

$$
u = x - \frac{9}{2}
$$

Then:

$$
du = dx
$$

Adjust the limits of integration:

- When $$ x = \frac{9}{2} $$, $$ u = 0 $$
- When $$ x = 5 $$, $$ u = 5 - \frac{9}{2} = \frac{1}{2} $$

Now the integral becomes:

$$
V_2 = \pi \int_{0}^{1/2} \left( \frac{1}{4} - u^2 \right) \, du
$$

### Evaluating the Integral

Separate the integral:

$$
V_2 = \pi \left[ \int_{0}^{1/2} \frac{1}{4} \, du - \int_{0}^{1/2} u^2 \, du \right]
$$

Compute each integral separately:

1. **First Integral:**

$$
\int_{0}^{1/2} \frac{1}{4} \, du = \frac{1}{4} \cdot u \Big|_{0}^{1/2} = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}
$$

2. **Second Integral:**

$$
\int_{0}^{1/2} u^2 \, du = \frac{u^3}{3} \Big|_{0}^{1/2} = \frac{\left(\frac{1}{2}\right)^3}{3} = \frac{1}{24}
$$

### Combine the Results

Subtract the second integral from the first:

$$
V_2 = \pi \left( \frac{1}{8} - \frac{1}{24} \right) = \pi \left( \frac{3}{24} - \frac{1}{24} \right) = \pi \cdot \frac{2}{24} = \pi \cdot \frac{1}{12}
$$

### Final Result

$$
V_2 = \frac{\pi}{12}
$$

So, the volume $$ V_2 $$ is $$ \frac{\pi}{12} $$ cubic units.
The volume $$ V_2 $$ is $$ \frac{\pi}{12} $$ cubic units.

\( \left. \begin{array} { l }{ V _ { 2 } = \pi \int _ { 9 / 2 } ^ { 5 } ( f ( x ) ^ { 2 } - g ( x ) ^ { 2 } ) d y } \\ { = \pi \int _ { 9 / 2 } ^ { 5 } ( ( 1 / 2 ) ^ { 2 } - ( x - 9 / 2 ) ^ { 2 } ) d x } \end{array} \right. \)

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