Find the area of the region bounded by the cu | vedclass

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

Question

(A) The given curve is $y=x^{2}$,which is a parabola symmetric about the $y$-axis with its vertex at the origin $(0,0)$.The region is bounded by the c

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) The given curve is $y=x^{2}$,which is a parabola symmetric about the $y$-axis with its vertex at the origin $(0,0)$.The region is bounded by the curve $y=x^{2}$ and the line $y=4$. The intersection points of the curve and the line are found by substituting $y=4$ into $y=x^{2}$,which gives $x^{2}=4$,so $x=\pm 2$.Since the parabola is symmetric about the $y$-axis,the total area $A$ is twice the area of the region in the first quadrant bounded by the curve,the $y$-axis,and the line $y=4$.Using horizontal strips of width $dy$ and length $x = \sqrt{y}$,the area is given by:$A = 2 \int_{0}^{4} x \, dy = 2 \int_{0}^{4} \sqrt{y} \, dy$Evaluating the integral:$A = 2 \left[ \frac{y^{3/2}}{3/2} ight]_{0}^{4} = 2 	imes \frac{2}{3} \left[ y^{3/2} ight]_{0}^{4}$$A = \frac{4}{3} \left( 4^{3/2} - 0^{3/2} ight) = \frac{4}{3} 	imes 8 = \frac{32}{3}$Thus,the area of the region is $\frac{32}{3}$ square units.

Find the area of the region bounded by the curve $y=x^{2}$ and the line $y=4$. (in $/3$)

Similar Questions

Area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$ is ......... $sq. \text{ unit}$.

If $x^{2}+y^{2}=a^{2}$, then $\int_{0}^{a} \sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx=$

Find the area of the region in the first quadrant enclosed by the $x-$ axis,the line $x=\sqrt{3} y$ and the circle $x^{2}+y^{2}=4$.

The area of the region bounded by the curve $y = x^2 - x - 6$ and the $X$-axis is . . . . . . sq. units.

The area (in sq. units) of the region bounded by the curve $y = 2x - x^2$ and the $X$-axis is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module