Question

Set up an integral to find the area of the surface generated by rotating the curve y = 4 ln x, from x = 1 to x = 3

(a) About the x-axis.
(b) About the line y = −3.
(c) About the line x = −3.

Answer 1

a)
`\(2\pi \int_(1)^(3) 4 \ln x \sqrt{1 + \left((1)/(x)\right)^2} \, dx\)`

b)
`\(2\pi \int_(1)^(3) (4 \ln x + 3) \sqrt{1 + \left((1)/(x)\right)^2} \, dx\)`

c)
`\(2\pi \int_(\ln(1))^(\ln(3)) e^(y/4) \sqrt{1 + \left((4)/(y)\right)^2} \, dy\)`

Step-by-step explanation:

a) About the x-axis:

The formula to find the surface area generated by rotating the curve
`\(y = 4 \ln x\)` from
`\(x = 1\) to \(x = 3\)` about the x-axis is given by the surface area integral:

`\[ A = 2\pi \int_(a)^(b) f(x) √(1 + [f'(x)]^2) \, dx \]`

Here,
`\(f(x) = 4 \ln x\), and \(f'(x)\) is the derivative of \(f(x)\)`.

`\[ A_x = 2\pi \int_(1)^(3) 4 \ln x \sqrt{1 + \left((1)/(x)\right)^2} \, dx \]`

b) About the line y = −3:

The formula for the surface area when rotating the curve about the line
`\(y = -3\)` is:

`\[ A = 2\pi \int_(a)^(b) [f(x) - (-3)] √(1 + [f'(x)]^2) \, dx \]`

`\[ A_y = 2\pi \int_(1)^(3) (4 \ln x + 3) \sqrt{1 + \left((1)/(x)\right)^2} \, dx \]`

c) About the line x = −3:

When rotating about the line
`\(x = -3\)`, the formula is:

`\[ A = 2\pi \int_(a)^(b) f(y) √(1 + [f'(y)]^2) \, dy \]`

Here,
`\(f(y)\)` is the inverse function of
`\(y = 4 \ln x\)`, which is
`\(x = e^(y/4)\)`. The derivative
`\(f'(y)\)` is found accordingly.

`\[ A_(x=-3) = 2\pi \int_(\ln(1))^(\ln(3)) e^(y/4) \sqrt{1 + \left((4)/(y)\right)^2} \, dy \]`

`\(x = e^(y/4)\)`

These integrals represent the surface area for rotation about the x-axis, y-axis, and a vertical line x = -3, respectively. Please note that the integrals themselves would need to be solved to find the numerical values.