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

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

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

$$0\leq y\leq 6$$

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

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

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

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

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

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