Find the exact area of the surface obtained by rotating the given curve about the x-axis. $x = 2 \cos ^ { 3 } \theta , \quad y = 2 \sin ^ { 3 } \theta , \quad 0 \leq \theta \leq \pi / 2$
A) $\frac { 24 \pi } { 5 }$
B) $\frac { 18 \pi } { 5 }$
C) $\frac { 12 \pi } { 5 }$
D) $\frac { 2 \pi } { 4 }$
E) None of these

Question

Quizplus · Accepted Answer

The surface area $S$ of a curve rotated about the x-axis from $a$ to $b$ is given by $S = 2\pi \int_{a}^{b} y \sqrt{1 + (\frac{dy}{dx})^2} dx$. For the given parametric equations $x = 2\cos^3\theta$ and $y = 2\sin^3\theta$, we first find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$, then express $\frac{dy}{dx}$ in terms of $\theta$, and finally integrate with respect to $\theta$ from 0 to $\pi/2$. The correct calculation leads to the area of $\frac{24\pi}{5}$.

Find the Exact Area of the Surface Obtained by Rotating x=2cos⁡3θ,y=2sin⁡3θ,0≤θ≤π/2x = 2 \cos ^ { 3 } \theta , \quad y = 2 \sin ^ { 3 } \theta , \quad 0 \leq \theta \leq \pi / 2x=2cos3θ,y=2sin3θ,0≤θ≤π/2

Find the Exact Area of the Surface Obtained by Rotating $x = 2 \cos ^ { 3 } \theta , \quad y = 2 \sin ^ { 3 } \theta , \quad 0 \leq \theta \leq \pi / 2$