Area of loop described as x = 3 t ( 6 − t ) , y = 8 t 2 ( 6 − t ) is -

0 ≤ t ≤ 6 , d t d x = 2 − 3 2 t . S = ∫ 0 6 y d t d x d t = 12 1 ∫ 0 6 t 2 ( 6 − t ) ( 3 − t ) d t = 12 1 ∫ 0 6 ( 18 t 2 − 9 t 3 + t 4 ) d t = 12 1 1296 − 2916 + 5 7776 = 5 27 .

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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by Miscellaneous CurvesMedium2 minQB

MathematicsMediumsingle choice

Area of loop described as $x = \frac{t}{3} (6 - t)$ , $y = \frac{t ^{2}}{8} (6 - t)$ is -

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Solution:

$0 \leq t \leq 6, \frac{d x}{d t} = 2 - \frac{2 t}{3} . S = \int_{0}^{6} y \frac{d x}{d t} d t = \frac{1}{12} \int_{0}^{6} t^{2} (6 - t) (3 - t) d t = \frac{1}{12} \int_{0}^{6} (18 t^{2} - 9 t^{3} + t^{4}) d t = \frac{1}{12} 1296 - 2916 + \frac{7776}{5} = \frac{27}{5} .$

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by Miscellaneous Curves

⏱ 2mℹ️ Source: QB

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