Answer to Question #205272 in Calculus for lawrence canda

Question

Find the surface area generated when the portion of the curve

y² = 12x from x = 0, x = 3 is rotated about the x-axis.

Answer 1

If the curve is described as $x=g(y), c\leq y\leq d,$ then the formula for surface area

becomes

$S=\displaystyle\int_{c}^{d}2\pi y\sqrt{1+(x'_y)^2}dy$

$x=\dfrac{1}{12}y^2=>x'_y=\dfrac{1}{6}y$

$0\leq y\leq 6$

$S=\displaystyle\int_{0}^{6}2\pi y\sqrt{1+(\dfrac{1}{6}y)^2}dy$

$u=1+\dfrac{1}{36}y^2, du=\dfrac{1}{18}ydy$

$\int y\sqrt{1+\dfrac{1}{36}y^2}\ dy=18\int\sqrt{u}du=12u^{3/2}+C$

$=12(1+\dfrac{1}{36}y^2)^{3/2}+C$

$S=24\pi\big[(1+\dfrac{1}{36}y^2)^{3/2}\big]\begin{matrix} 6 \\ 0 \end{matrix}=24\pi(2\sqrt{2}-1) (units^2)$

$S=24\pi(2\sqrt{2}-1) \text{ square units}$