The area of the region bounded by the curve y | vedclass

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

Question

(B) The given curve is $y=x^{2}$ and the line is $y=16$.To find the area,we integrate with respect to $y$ from $y=0$ to $y=16$.The curve $y=x^{2}$ imp

Vedclass · Accepted Answer

$ \frac{256}{3} $ sq. units

Vedclass · Answer

(B) The given curve is $y=x^{2}$ and the line is $y=16$.To find the area,we integrate with respect to $y$ from $y=0$ to $y=16$.The curve $y=x^{2}$ implies $x = \pm \sqrt{y}$.The area is given by the integral:$Area = \int_{-4}^{4} (16 - x^{2}) dx$$= 2 \int_{0}^{4} (16 - x^{2}) dx$$= 2 [16x - \frac{x^{3}}{3}]_{0}^{4}$$= 2 [16(4) - \frac{4^{3}}{3}]$$= 2 [64 - \frac{64}{3}]$$= 2 [\frac{192 - 64}{3}]$$= 2 [\frac{128}{3}] = \frac{256}{3} 	ext{ sq. units}$.Alternatively,using integration with respect to $y$:$Area = 2 \int_{0}^{16} \sqrt{y} dy$$= 2 [\frac{y^{3/2}}{3/2}]_{0}^{16}$$= 2 	imes \frac{2}{3} [y^{3/2}]_{0}^{16}$$= \frac{4}{3} [16^{3/2}]$$= \frac{4}{3} [64] = \frac{256}{3} 	ext{ sq. units}$.

The area of the region bounded by the curve $y=x^{2}$ and the line $y=16$ is

Similar Questions

The increasing order of stability of the following free radicals is:

Which of the following is the correct order of size of the given species?

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

Which of the following complexes is not a chelate?

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod perpendicular to its length. The acceleration of the centre of mass of the rod is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module