The area bounded by the curves $y^{2} = 12 x$ and $x^{2} = 12 y$ is divided by the line $x = 3$ in two parts. The area (in square units) of the larger part is:

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$\frac{147}{4}$

$\frac{45}{4}$

$\frac{137}{4}$

$\frac{245}{4}$

answer is A.

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Detailed Solution

Reqd. area

$\begin{array}{l} = \int_{3}^{12} (\sqrt{12 x} - \frac{x^{2}}{12}) d x \\ = {\frac{2}{3} \sqrt{12} x^{3 / 2}|}_{3}^{12} - \frac{1}{36} [12^{3} - 3^{3}] \\ = 84 - \frac{567}{12} = \frac{441}{12} = \frac{147}{4} . \end{array}$

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