The given arc is $y = 1 - \dfrac {x^2}{4}$ about the $y$-axis from 0 to 2. Here is the farthest part I could ever go through.
$$y’ = -\frac{x}{2}$$
$$[y’]^2 = \frac{x^2}{4}$$
So $$\int_0^2x\cdot\sqrt{1+\frac{x^2}{4}}\,dx$$ What should I do next? Thanks for those who will help. I have a quiz later and I can’t solve similar problems like this.
Since you are trying to find the surface area for this surface of revolution which is revolved around the $y$-axis, the formula to compute it is $$A = \int_a^b 2 \pi x \sqrt{1+(f'(x))^2}dx$$ Note in your work, you are missing the $2\pi$. To compute your integral, you need a $u$-substitution. Try $u= 1+\frac{x^2}{4}$. Then $du = \frac{x}{2}dx \Rightarrow dx=\frac{2du}{x}$. Your bounds will also change, since you are now doing an integral with respect to $u$ instead of $x$. $$\int_0^2 \to \int_1^2$$ These new bounds come from plugging in $0$ and $2$ into your $u=1+\frac{x^2}{4}$ equation. Your new integral looks like: $$A = \int_1^2 2\pi x \sqrt{u}\frac{2du}{x} = 4 \pi \int_1^2 \sqrt{u}du$$
And from here, it is an integral that you can compute. Hope this helps!
