[Math] Find the exact area of the surface obtained by rotating the curve about the x-axis. x = 2 + 3y2, 1 ≤ y ≤ 2

Question

x = 2+3y^2

x' = 6y

$$\int_b^a y\sqrt{1+(x')^2}$$

$$\int_1^2 \sqrt{1+(6y)^2} \,dx= \frac\pi{36} (144\sqrt{144} – 37\sqrt{37})$$

Hi,

This is the work I have done to figure out the exact area of the surface obtained by rotating by the curve about the x-axis, but the website I use for my homework says that I am incorrect.

Please tell me if I did something wrong.

Thanks!!!

Answer 1

The formula is $2\pi \int_{a}^b y \sqrt{1+ (x')^2} dy = 2\pi \int_{1}^2 y \sqrt{1+ (6y)^2} dy$ = $\frac{1}{54}(145 \sqrt{145}-37 \sqrt{37}) \pi$.

1. In the formula, don't forget to multiply $2\pi$.

2. In the second integral, it is $dy$ instead of $dx$.