B. $$ \int_{0}^{1}\left(2 \pi(16-x) x^{2}\right) d x $$ The volume of the solid is $$ \square $$. (Type an exact answer.)

Question

UpStudy AI · Accepted Answer

We start with the integral

$$
\int_{0}^{1} 2\pi (16-x)x^2\,dx.
$$

**Step 1. Factor out the constant:**

$$
2\pi \int_{0}^{1} (16-x)x^2\,dx.
$$

**Step 2. Expand the integrand:**

$$
(16-x)x^2 = 16x^2 - x^3.
$$

Thus, the integral becomes

$$
2\pi \int_{0}^{1} \left(16x^2 - x^3\right) dx.
$$

**Step 3. Integrate term by term:**

$$
\int_{0}^{1} 16x^2\,dx = 16 \int_{0}^{1}x^2\,dx = 16\left[\frac{x^3}{3}\right]_{0}^{1} = 16\left(\frac{1}{3}\right)=\frac{16}{3}.
$$

$$
\int_{0}^{1} x^3\,dx = \left[\frac{x^4}{4}\right]_{0}^{1} = \frac{1}{4}.
$$

Subtracting the results:

$$
\int_{0}^{1} \left(16x^2 - x^3\right) dx = \frac{16}{3} - \frac{1}{4}.
$$

**Step 4. Combine the terms using a common denominator:**

$$
\frac{16}{3} - \frac{1}{4} = \frac{64}{12} - \frac{3}{12} = \frac{61}{12}.
$$

**Step 5. Multiply by $$2\pi$$:**

$$
2\pi \cdot \frac{61}{12} = \frac{122\pi}{12} = \frac{61\pi}{6}.
$$

Thus, the volume of the solid is

$$
\frac{61\pi}{6}.
$$
The volume of the solid is $$ \frac{61\pi}{6} $$.

B.
The volume of the solid is .
(Type an exact answer.)