Uppgift 1

Vi använder den högra parabelgrenen $x=\sqrt{24y}$, $0\le y\le18$. Rotation kring $y$-axeln ger samma yta som om båda grenarna tas med. Eftersom
$$\frac{dx}{dy}=\frac{12}{x}$$
blir rotationsytans area
$$\begin{aligned} A&=2\pi\int_0^{18}x\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy =2\pi\int_0^{18}\sqrt{24y+144}\,dy\\ &=\frac{\pi}{18}\left[(24y+144)^{3/2}\right]_0^{18} =\frac{\pi}{18}\left(576^{3/2}-144^{3/2}\right) =672\pi. \end{aligned}$$