Answer to Question #205269 in Calculus for ian dizon

Question

Find the surface area of the portion of the curve x² =4y from y =1 to y =2 when it is revolved about the y-axis.

Answer 1

$x^2=4y\quad \Rightarrow \quad x=\sqrt{4y}=2\sqrt{y}$

$g(y)=2\sqrt{y}$ from $y=1$ to $y=2$

$g^\prime (y)=1/\sqrt{y}$

Surface Area $=\int\limits_1^22\pi \ g(y)\ \sqrt{1+\big(g^\prime (y)\big)^2}\ dy=\int\limits_1^24\pi \sqrt{y} \ \cdot \ \sqrt{1+1/y}\ dy=4\pi \int\limits_1^2\sqrt{y+1}\ dy= \tfrac{8\pi}{3} (y+1)^{3/2}\bigg|_1^2=\tfrac{8\pi}{3}\big(3\sqrt{3}-2\sqrt{2}\big)$

Answer: Surface Area $\approx \ 19.836$