Area of a surface of revolution $x=e^{2y}$ between $y=0$ and $y=\frac12$

Question

I have the following function: $x=e^{2y}$ between $y=0$ and $y=\frac12$.

I try to calculate the surface area of revolution after turning the function around the $y$-axis.

The integral that I have left is: $$\int_0^{\frac12}2\pi e^{2y}\sqrt{1+4e^{4y}}\,dy$$

Answer 1

Hint: The integral can be written as$$\int_0^\frac12 2\pi e^{2y}\sqrt{1+4e^{4y}}\,dy=\int_1^e \pi \sqrt{1+4u^2}\,du$$ using the substitution $u=e^{2y}$.

Answer 2

Hint:

Use substitution $2e^{2y}=u$ to find $\displaystyle\pi\int_{2}^{2e}\sqrt{1+u^2}du$ and now let $u=\sinh u$.