Find the area bounded on the left by the y-ax | vedclass

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

Question

(B) The region is bounded by $x=0$ (the $y-$axis) on the left,$x=\frac{\pi}{2}$ on the right,and the $x-$axis $(y=0)$ below.The upper boundary is defi

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(B) The region is bounded by $x=0$ (the $y-$axis) on the left,$x=\frac{\pi}{2}$ on the right,and the $x-$axis $(y=0)$ below.The upper boundary is defined by $y = \cos x$ for $0 \le x \le \frac{\pi}{4}$ and $y = \sin x$ for $\frac{\pi}{4} \le x \le \frac{\pi}{2}$,since $\cos x = \sin x$ at $x = \frac{\pi}{4}$.The required area $A$ is given by:$A = \int_{0}^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi/2} \sin x \, dx$$A = [\sin x]_{0}^{\pi/4} + [-\cos x]_{\pi/4}^{\pi/2}$$A = (\sin(\frac{\pi}{4}) - \sin(0)) + (-(\cos(\frac{\pi}{2}) - \cos(\frac{\pi}{4})))$$A = (\frac{1}{\sqrt{2}} - 0) + (-(0 - \frac{1}{\sqrt{2}}))$$A = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Find the area bounded on the left by the $y-$axis,below by the $x-$axis,on the right by $x = \frac{\pi}{2}$,above on the left by $y = \cos x$,and above on the right by $y = \sin x$.

Similar Questions

The area bounded by the parabolas $y=4x^{2}$, $y=\frac{x^{2}}{9}$ and the straight line $y=2$ is

The area of the region enclosed by $y \leq 4x^{2}$,$x^{2} \leq 9y$ and $y \leq 4$ is equal to:

The area of the region $A = \{(x, y): 4x^2 + y^2 \le 8 \text{ and } y^2 \le 4x\}$ is:

The area of the region bounded by $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is

The volume of the solid generated by revolving about the $y-$axis the figure bounded by the parabolas $y = x^2$ and $x = y^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module