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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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

Similar Questions

One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,where $\ell, m, n \in N$. Then $\ell+m+n$ is equal to

The area of the region $R=\{(x,y):xy \le 8, 1 \le y \le x^{2}, x \ge 0\}$ is

The area bounded by the curve $y^2=2x+1$ and the line $x-y=1$ is

The area of the figure bounded by the parabolas $x = -2y^{2}$ and $x = 1 - 3y^{2}$ is

The area bounded by the curves $y-1=\cos x$,$y=\sin x$ and the $X$-axis between $x=0$ and $x=\pi$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module