The area of the loop of the curve x = a ( 1 − t 2 ) , y = a t ( 1 − t 2 ) , − 1 ≤ t ≤ 1 must be

As t goes from -1 to 1 the curve starts and ends at the origin, describing one loop. A = − ∫ − 1 1 g ( t ) f ′ ( t ) d t = − ∫ − 1 1 a t ( 1 − t 2 ) ( − 2 a t ) d t = 2 a 2 ∫ − 1 1 ( t 2 − t 4 ) d t = 4 a 2 ( 3 1 − 5 1 ) = 15 8 a 2 .

The area of the loop of the curve x = a ( 1 − t 2 ) , y = a t ( 1 − t 2 ) , − 1 | 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 the following questions.

The area of the loop of the curve $x = a (1 - t^{2}), y = a t (1 - t^{2}), - 1 \leq t \leq 1 must be$

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Solution:

As $t$ goes from -1 to 1 the curve starts and ends at the origin, describing one loop. $A =$ $- \int_{- 1}^{1} g (t) f^{'} (t) d t = - \int_{- 1}^{1} a t (1 - t^{2}) (- 2 a t) d t = 2 a^{2} \int_{- 1}^{1} (t^{2} - t^{4}) d t = 4 a^{2} (\frac{1}{3} - \frac{1}{5}) = \frac{8 a ^{2}}{15}$ .

Mathematics - Area Under Curves Question with Solution | TestHub

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Doubts & Discussion