The area bounded by the curves {y^2} = 8x and | vedclass

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

Question

(B) The given curves are ${y^2} = 8x$ and $y = x$. To find the points of intersection,substitute $y = x$ into ${y^2} = 8x$: ${x^2} = 8x \Rightarrow {x

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given curves are ${y^2} = 8x$ and $y = x$. To find the points of intersection,substitute $y = x$ into ${y^2} = 8x$: ${x^2} = 8x \Rightarrow {x^2} - 8x = 0 \Rightarrow x(x - 8) = 0$. Thus,the points of intersection are $x = 0$ and $x = 8$. The required area is given by the integral: $	ext{Area} = \int_{0}^{8} (\sqrt{8x} - x) \, dx = \int_{0}^{8} (2\sqrt{2}x^{1/2} - x) \, dx$. Evaluating the integral: $= \left[ 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} ight]_{0}^{8} = \left[ \frac{4\sqrt{2}}{3} x^{3/2} - \frac{x^2}{2} ight]_{0}^{8}$. Substituting the limits: $= \left( \frac{4\sqrt{2}}{3} (8)^{3/2} - \frac{8^2}{2} ight) - 0 = \left( \frac{4\sqrt{2}}{3} \cdot 16\sqrt{2} - 32 ight) = \frac{128}{3} - 32 = \frac{128 - 96}{3} = \frac{32}{3} 	ext{ sq. units}$.

The area bounded by the curves ${y^2} = 8x$ and $y = x$ is

Similar Questions

The area of the region (in sq. units) bounded by the curves $x^2+y^2=16$ and $y^2=6x$ is

Find the area enclosed by the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$.

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

The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2x$ is equal to:

Area of the region enclosed between the curves $y^2=4(x+7)$ and $y^2=5(2-x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module