Question

find the surface area of the portion of the curve x^2+y^2=4 from x=0 to x=2 when it is revolved about the y-axis

Answer 1

$$S=2\pi\displaystyle\int_{a}^{b}r\cos (t)\sqrt{(r(t))^2+(r'(t))^2}dt$$

$$x^2+y^2=4$$

$$r=2, a=-\pi/2, b=\pi/2$$

$$S=2\pi\displaystyle\int_{-\pi/2}^{\pi/2}2\cos t\sqrt{(2)^2+(0)^2}dt$$

$$=8\pi[\sin t]\begin{matrix} \pi/2 \\ -\pi/2 \end{matrix}=16\pi({units}^2)$$