Question

Find the exact area of the surface obtained by rotating the curve about the x-axis.
y =

1 + 3x
, 1 ≤ x ≤ 3

Answer 1

To find the exact area of the surface obtained by rotating the curve \(y = 1 + 3x\) about the x-axis from \(x = 1\) to \(x = 3\), we use the formula for the surface area of revolution about the x-axis:

\[
S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
\]

1. **Find \(\frac{dy}{dx}\)**:
The derivative of \(y\) with respect to \(x\) is:
\[
\frac{dy}{dx} = 3
\]

2. **Set up the integral**:
Now we need to compute \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\):
\[
\left(\frac{dy}{dx}\right)^2 = 3^2 = 9
\]
Therefore:
\[
\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

3. **Plug everything into the surface area formula**:
Substituting \(y = 1 + 3x\) and \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{10}\) into the integral, we get:
\[
S = 2\pi \int_1^3 (1 + 3x) \sqrt{10} \, dx
\]
This can be simplified to:
\[
S = 2\pi \sqrt{10} \int_1^3 (1 + 3x) \, dx
\]

4. **Evaluate the integral**:
We can split the integral into two parts:
\[
\int_1^3 (1 + 3x) \, dx = \int_1^3 1 \, dx + \int_1^3 3x \, dx
\]

- The first integral:
\[
\int_1^3 1 \, dx = [x]_1^3 = 3 - 1 = 2
\]

- The second integral:
\[
\int_1^3 3x \, dx = 3 \left[\frac{x^2}{2}\right]_1^3 = 3 \left[\frac{9}{2} - \frac{1}{2}\right] = 3 \cdot \frac{8}{2} = 12
\]

So, we have:
\[
\int_1^3 (1 + 3x) \, dx = 2 + 12 = 14
\]

5. **Substituting back into the surface area formula**:
\[
S = 2\pi \sqrt{10} (14) = 28\pi \sqrt{10}
\]

Thus, the exact area of the surface obtained by rotating the curve about the x-axis is:

\[
\boxed{28\pi \sqrt{10}}
\]