Area of the region bounded by the curve y^{2} | vedclass

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

Question

(B) The area bounded by the curve $y^{2}=4x$,the $y$-axis,and the line $y=3$ is calculated by integrating $x$ with respect to $y$ from $y=0$ to $y=3$.

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(B) The area bounded by the curve $y^{2}=4x$,the $y$-axis,and the line $y=3$ is calculated by integrating $x$ with respect to $y$ from $y=0$ to $y=3$.Given the curve $y^{2}=4x$,we can express $x$ as $x = \frac{y^{2}}{4}$.The area $A$ is given by the integral:$A = \int_{0}^{3} x \, dy$Substituting $x = \frac{y^{2}}{4}$:$A = \int_{0}^{3} \frac{y^{2}}{4} \, dy$$A = \frac{1}{4} \left[ \frac{y^{3}}{3} ight]_{0}^{3}$$A = \frac{1}{4} \left( \frac{3^{3}}{3} - \frac{0^{3}}{3} ight)$$A = \frac{1}{4} \left( \frac{27}{3} ight)$$A = \frac{1}{4} (9) = \frac{9}{4} 	ext{ sq. units}$.Thus,the correct answer is $B$.

Area of the region bounded by the curve $y^{2}=4x$,$y$-axis and the line $y=3$ is

Similar Questions

The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is $24$ sq. units. Then,$a = $ . . . . . . .

The area enclosed by the curve $y^2 + x^4 = x^2$ is :

The area (in sq units) between the curve $y^2 = 8x$ and its latus rectum is

$A$ frustum of a sphere is made by cutting a sphere with two parallel planes. If the radius of the sphere is $5 \, cm$ and the distance between the planes is $1 \, cm$,what is the curved surface area of the frustum when the distance of the first plane from the center of the sphere is $2 \, cm$?

Find the area bounded by the curves $\{(x, y): y \geq x^{2} \text{ and } y=|x|\}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module