Answer to Question #346756 in Calculus for edz

Question

compute the surface area generated when the first quadrant portion of the curve x^2-4y=8 from x1=0 to x2= 2 is revolved about the y-axis

Answer 1

$y=\dfrac{x^2}{4}-2$

$y'=\dfrac{x}{2}$

$S=2\pi\displaystyle\int_{0}^{2}x\sqrt{(y')^2+1}dx$

$=2\pi\displaystyle\int_{0}^{2}x\sqrt{(\dfrac{x}{2})^2+1}dx$

$\int x\sqrt{\dfrac{x^2}{4}+1}dx$

$u=\dfrac{x^2}{4}+1, du=\dfrac{x}{2}dx$

$\int x\sqrt{\dfrac{x^2}{4}+1}dx=\int 2\sqrt{u}du$

$=2(\dfrac{2}{3})u^{3/2}+C=\dfrac{4}{3}(\dfrac{x^2}{4}+1)^{3/2}+C$

$S=2\pi\displaystyle\int_{0}^{2}x\sqrt{(y')^2+1}dx$

$=2\pi\displaystyle\int_{0}^{2}x\sqrt{(\dfrac{x}{2})^2+1}dx$

$=2\pi[\dfrac{4}{3}(\dfrac{x^2}{4}+1)^{3/2}]\begin{matrix} 2\\ 0 \end{matrix}$

$=\dfrac{8\pi}{3}(2\sqrt{2}-1)({units}^2)$