Area enclosed by the ellipse $9x^2 + 4y^2 = 1$ in the first quadrant is . . . . . . .

Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$.

Let $g(x) = \cos(x^2)$,$f(x) = \sqrt{x}$,and $\alpha, \beta$ (where $\alpha < \beta$) be the roots of the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Then,the area (in sq. units) bounded by the curve $y = (g \circ f)(x)$ and the lines $x = \alpha$,$x = \beta$,and $y = 0$ is:

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

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

Find the area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$.