The area bounded between the curve x^2=y and | vedclass

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

Question

(A) To find the area bounded between the curve $x^2=y$ and the line $y=4x$,we first find their points of intersection.Setting $x^2 = 4x$,we get $x^2 -

Vedclass · Accepted Answer

$\frac{32}{3}$ sq. units

Vedclass · Answer

(A) To find the area bounded between the curve $x^2=y$ and the line $y=4x$,we first find their points of intersection.Setting $x^2 = 4x$,we get $x^2 - 4x = 0$,which implies $x(x-4) = 0$. Thus,$x=0$ and $x=4$.For $x=0$,$y=0$,and for $x=4$,$y=16$. So the points of intersection are $(0,0)$ and $(4,16)$.The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:$A = \int_{0}^{4} (4x - x^2) dx$$A = \left[ \frac{4x^2}{2} - \frac{x^3}{3} ight]_{0}^{4}$$A = \left[ 2x^2 - \frac{x^3}{3} ight]_{0}^{4}$$A = \left( 2(4)^2 - \frac{(4)^3}{3} ight) - (0 - 0)$$A = \left( 32 - \frac{64}{3} ight)$$A = \frac{96 - 64}{3} = \frac{32}{3} 	ext{ sq. units}$

The area bounded between the curve $x^2=y$ and the line $y=4x$ is

Similar Questions

Find the area of the circle $4x^{2}+4y^{2}=9$ which is interior to the parabola $x^{2}=4y$.

The area of the region $\{(x, y):|x-y| \leq y \leq 4 \sqrt{x}\}$ is

The area bounded by the curve $y^2 = x$,the line $y = 4$,and the $y$-axis is

The volume of the solid generated by revolving the region bounded by the parabolas $y=x^{2}$ and $x=y^{2}$ about the $y$-axis is:

The area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $x = 1$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module