<s>[INST] How to find surface area of $x=\sqrt{a^2-y^2}$ I still hard time to find surface area of function...
I have 

The given curve is rotated about the $y$-axis. Find the area of the resulting surface. $$x= \sqrt{a^2-y^2},\quad y\in\left[0,\frac{a}{2}\right]$$

If you have any idea, would you post your idea ? Thank you :)
 [/INST] We have formula 
$$\text{Surface area } S = \int 2\pi x \sqrt{1+\left(\frac{dx}{dy}\right)^2} dy$$
we have $x= \sqrt{a^2-y^2}$ and  $\frac{dx}{dy}= \frac{-y}{\sqrt{a^2-y^2}}$. Hence we have,
$$S=\int_0^{\frac{a}{2}}2\pi \sqrt{a^2-y^2}\sqrt{1+ \frac{y^2}{{a^2-y^2}}} dy$$
$$=\int_0^{\frac{a}{2}}2\pi \sqrt{a^2-y^2} \frac{a}{\sqrt{a^2-y^2}}dy $$
$$=\int_0^{\frac{a}{2}}2\pi .a dy$$
$$=\pi .a^2$$
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