Question

Apply Green's theorem to find the area of the ellipse (x2)/9+(y2)/4=4 .

Answer 1

Answer on Question#41421 – Math - Integral Calculus

We have the equation of the ellipse:

$$\left. \frac {x ^ {2}}{9} + \frac {y ^ {2}}{4} = 4 \right| \times \frac {1}{4}$$

or

$$\frac {x ^ {2}}{4 \cdot 3 ^ {2}} + \frac {y ^ {2}}{4 \cdot 2 ^ {2}} = 1$$

or

$$\frac {x ^ {2}}{6 ^ {2}} + \frac {y ^ {2}}{4 ^ {2}} = 1$$

Let $A$ be the area of the region $D$ bounded by the ellipse. Since we may parameterize $\partial D$, with counterclockwise orientation, by $\varphi(t) = (6\cos t, 4\sin t)$,

where $t \in [0,2\pi]$. Then using Green's theorem we have

$$\begin{array}{l} A = \frac {1}{2} \int_ {\partial D} x d y - y d x = \frac {1}{2} \int_ {0} ^ {2 \pi} (- 4 \sin t, 6 \cos t) \cdot (- 6 \sin t, 4 \cos t) d t \\ = \frac {1}{2} \int_ {0} ^ {2 \pi} (6 \cdot 4 \sin^ {2} t + 6 \cdot 4 \cos^ {2} t) d t = \frac {6 \cdot 4}{2} \int_ {0} ^ {2 \pi} d t = 12 \cdot 2 \pi = 24 \pi \end{array}$$

Answer: $A = 24\pi$.

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