The area (in sq. units) bounded by the curves | vedclass

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(C) The given curves are $x^2=9y$ and $(x-6)^2=9y$. To find the intersection point,set $9y = 9y$: $x^2 = (x-6)^2$ $x^2 = x^2 - 12x + 36$ $12x = 36 \im

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$2$

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(C) The given curves are $x^2=9y$ and $(x-6)^2=9y$. To find the intersection point,set $9y = 9y$: $x^2 = (x-6)^2$ $x^2 = x^2 - 12x + 36$ $12x = 36 \implies x = 3$. At $x=3$,$y = \frac{3^2}{9} = 1$. So,the intersection point is $(3, 1)$. The area bounded by the curves and the $X$-axis is the sum of the areas under the two parabolas from $x=0$ to $x=3$ and $x=3$ to $x=6$ respectively: $	ext{Required Area} = \int_0^3 \frac{x^2}{9} dx + \int_3^6 \frac{(x-6)^2}{9} dx$ $= \frac{1}{9} \left[ \frac{x^3}{3} ight]_0^3 + \frac{1}{9} \left[ \frac{(x-6)^3}{3} ight]_3^6$ $= \frac{1}{27} [3^3 - 0^3] + \frac{1}{27} [(6-6)^3 - (3-6)^3]$ $= \frac{27}{27} + \frac{1}{27} [0 - (-27)]$ $= 1 + \frac{27}{27} = 1 + 1 = 2 	ext{ sq. units.}$

The area (in sq. units) bounded by the curves $x^2=9y$,$(x-6)^2=9y$ and the $X$-axis is

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