Answer to Question #317915 in Calculus for jinxh

Question

find the area of the surface generated when the given arc is revolved about the y axis ( y= 4 - x^2 from x=0 to x=2 )

Answer 1

$x=\sqrt{4-y}\\S=2\pi \int_0^4{x\left( y \right) \sqrt{1+x'\left( y \right) ^2}dy}=2\pi \int_0^4{\sqrt{\left( 4-y \right) \left( 1+\left( -\frac{1}{2\sqrt{4-y}} \right) ^2 \right)}dy}=\\=2\pi \int_0^4{\sqrt{\frac{17}{4}-y}dy}=-2\pi \cdot \frac{\left( \frac{17}{4}-y \right) ^{3/2}}{3/2}|_{0}^{4}=\frac{\pi}{6}\left( 17^{3/2}-1 \right)$