If x^{2}+y^{2}=a^{2}, then int_{0}^{a} sqrt{1 | vedclass

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

Question

(C) Given the equation of the circle $x^{2}+y^{2}=a^{2}$.Differentiating with respect to $x$, we get $2x + 2y \frac{dy}{dx} = 0$, which implies $\frac

Vedclass · Accepted Answer

$\frac{1}{2} \pi a$

Vedclass · Answer

(C) Given the equation of the circle $x^{2}+y^{2}=a^{2}$.Differentiating with respect to $x$, we get $2x + 2y \frac{dy}{dx} = 0$, which implies $\frac{dy}{dx} = -\frac{x}{y}$.Now, substitute this into the integral: $\sqrt{1+\left(\frac{dy}{dx}\right)^{2}} = \sqrt{1+\frac{x^{2}}{y^{2}}} = \sqrt{\frac{y^{2}+x^{2}}{y^{2}}} = \sqrt{\frac{a^{2}}{y^{2}}} = \frac{a}{y}$.The integral becomes $\int_{0}^{a} \frac{a}{y} dx$.Since $y = \sqrt{a^{2}-x^{2}}$, the integral is $\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx$.Evaluating this, we get $a \left[ \sin^{-1} \left( \frac{x}{a} \right) \right]_{0}^{a} = a \left( \sin^{-1}(1) - \sin^{-1}(0) \right) = a \left( \frac{\pi}{2} - 0 \right) = \frac{\pi a}{2}$.

If $x^{2}+y^{2}=a^{2}$, then $\int_{0}^{a} \sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx=$

Similar Questions

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$ and $x = 2a$ has the least value,is

The area bounded by the curve $y = k \sin x$ between $x = \pi$ and $x = 2\pi$ is:

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

The area (in square units) bounded by the line $y = x$, the $X$-axis and the lines $x = -2$ and $x = 4$ is...

Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$,and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then $b$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module