The area of the region (in sq. units),in the | vedclass

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

Question

(D) The given parabola is $y = 9x^2$,which implies $x^2 = y/9$. Since we are in the first quadrant,$x = \sqrt{y}/3$.The area $A$ bounded by the curve

Vedclass · Accepted Answer

$14/9$

Vedclass · Answer

(D) The given parabola is $y = 9x^2$,which implies $x^2 = y/9$. Since we are in the first quadrant,$x = \sqrt{y}/3$.The area $A$ bounded by the curve $x = f(y)$,the $y$-axis $(x = 0)$,and the lines $y = 1$ and $y = 4$ is given by the integral:$A = \int_{1}^{4} x \, dy = \int_{1}^{4} \frac{\sqrt{y}}{3} \, dy$$A = \frac{1}{3} \int_{1}^{4} y^{1/2} \, dy$Using the power rule $\int y^n \, dy = \frac{y^{n+1}}{n+1}$,we get:$A = \frac{1}{3} \left[ \frac{y^{3/2}}{3/2} ight]_{1}^{4} = \frac{1}{3} 	imes \frac{2}{3} \left[ y^{3/2} ight]_{1}^{4}$$A = \frac{2}{9} \left( 4^{3/2} - 1^{3/2} ight)$$A = \frac{2}{9} (8 - 1) = \frac{2}{9} 	imes 7 = \frac{14}{9} 	ext{ sq. units.}$

The area of the region (in sq. units),in the first quadrant bounded by the parabola $y = 9x^2$ and the lines $x = 0, y = 1$ and $y = 4$,is

Similar Questions

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

The area of the smaller part between the circle $x^2+y^2=4$ and the line $x=1$ is . . . . . . sq. units.

The area of the region bounded by the curve $y^2=9x$ and the line $y=3x$ is

In the figure,$AOBA$ is the part of the ellipse $9x^{2} + y^{2} = 36$ in the first quadrant such that $OA = 2$ and $OB = 6$. Find the area between the arc $AB$ and the chord $AB$.

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