Answer to Question #98857 in Calculus for Rachel

Question

Use Green’s Theorem to find the area inside x^2/3+y^2/3= 1.

Answer 1

Green's Theorem

We need to find the Area inside ${x^\frac {2} {3}} + {y^\frac {2} {3}} = 1$

Solution:

Given,

${x^\frac {2} {3}} + {y^\frac {2} {3}} = 1$

Let

$x = cos^{ 3} \space {t} \space \space \space and \space y = sin^ {3} \space {t} , \space 0\le t \le 2\pi$

By Green's theorem

Area inside =

$\frac {1}{2} \oint_A ( x \space dy - y \space dx)$

We know,

$x = cos^{3} \space {t} \space$

$dx = {3} \space cos^{2} \space {(t)} \times (-sin (t)) = - 3 cos^2 (t) sin (t) \space dt$

$y = sin^ {3} \space {t}$

$dy = 3 sin^2 (t) \times cos (t) \space dt$

Now plug all these in the formula of Green's theorem

 Area inside  = $\frac {1}{2} \oint_A (( x \space dy - y \space dx) = \frac {1}{2} [\int_{t =0} ^ {2\pi} (cos^{ 3} \space {t} 3 sin^2 (t) \times cos (t) \space dt) - sin^ {3} \space {t} (- 3 cos^2 (t) sin (t)) \space dt)]$

$= \frac {3}{2} [\int_{t =0} ^ {2\pi} (cos^{ 4} \space {t} \space sin^2 (t) \times \space dt) + sin^ {4} \space {t} \space cos^2 (t) \space dt)]$

$= \frac {3}{2} [\int_{t =0} ^ {2\pi} cos^{ 2} \space {t} \space sin^2 (t) \times \space (cos^2 t + sin^ {2} \space {t} ) ] dt$

$= \frac {3}{2} \int_{t =0} ^ {2\pi} cos^{ 2} \space {t} \space sin^2 (t) dt$

$= \frac {3}{2} \times 2 \times 2 \int_{t =0} ^ {\frac {\pi} {2}} cos^{ 2} \space {t} \space sin^2 (t) dt$

$= 6 \int_{t =0} ^ {\frac {\pi} {2}} cos^{ 2} \space {t} \space sin^2 (t) dt$

$= \frac {6} {4} \int_{t =0} ^ {\frac {\pi} {2}} ( 2 sin t cos t)^2 dt = = \frac {6} {4} \int_{t =0} ^ {\frac {\pi} {2}} sin ^2 \space 2t \space dt$

( Since sin 2t = 2 sint t cos t)

$= \frac {6} {4} \int_{t =0} ^ {\frac {\pi} {2}} \frac { 1- cos 4t} {2} \space dt = \frac {3}{4} [ t - \frac {sin 4t}{4}]_{t =0} ^{ t = \frac {\pi}{2}}$

$= \frac {3}{4} \times \frac {\pi}{2} = \frac {3 \pi} {8}$

Answer: The Area inside = $\frac {3 \pi} {8}$