The area enclosed by the curve x = a cos 3 t , y = a sin 3 t , 0 ≤ t ≤ 2 π is | TestHub

Passage / Comprehension

If a curve is given by its parametric equations $x = f (t), y = g (t)$ and suppose the derivatives $f^{'} (t)$ and $g^{'} (t)$ are continuous functions on the interval $[t_{1}, t_{2}]$ . If $t_{1}$ and $t_{2}$ are the values of parameter $t$ corresponding respectively to the initial and final position in which the curve can be described as a contour in the positive direction (i.e., figure remains left) then the area described by the curve is $A = - \int_{t_{1}}^{t_{2}} g (t) f^{'} (t) d t = \int_{t_{1}}^{t_{2}} f (t) g^{'} (t) d t = \frac{1}{2} \int_{t_{1}}^{t_{2}} (x g^{'} (t) - y f^{'} (t)) d t$ . Answer t

The area enclosed by the curve $x = a cos^{3} t, y = a sin^{3} t, 0 \leq t \leq 2 π is$

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Solution:

$x g^{'} - y f^{'} = a cos^{3} t \cdot 3 a sin^{2} t cos t + a sin^{3} t \cdot 3 a cos^{2} t sin t = 3 a^{2} sin^{2} t cos^{2} t$ . Area $= \frac{1}{2} \int_{0}^{2 π} 3 a^{2} sin^{2} t cos^{2} t d t =$ $\frac{3 a ^{2}}{8} \int_{0}^{2 π} sin^{2} 2 t d t = \frac{3 a ^{2}}{8} \cdot π = \frac{3}{8} π a^{2}$ .

Mathematics - Area Under Curves Question with Solution | TestHub

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