91影视

The equation is$x^2+y^2=1$ and line $x=\frac{1}{2}$

The goal of this issue is to calculate the size of the region that lies inside the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=1$and to the right of the vertical line $\text{x=}\frac{1}{2}$

Draw a circle ${\mathrm{x}}^2+{\mathrm{y}}^2=1$and a line on a piece of paper on $\text{x=}\frac{1}{2}$

Graph of ${\mathrm{x}}^2+{\mathrm{y}}^2=1\text{and}\mathrm{x}=\frac{1}{2}$

Area of the circle's inner region $x^2+y^2=1$and the right-hand side of the vertical line $x=\frac{1}{2}$is ABCD.

The required area can be expressed as

$A={鈭�}_{x=1/2}^{x=1}鈥�{鈭�}_{y=鈭�\sqrt{1鈭�x^2}}^{y=\sqrt{1鈭�x^2}}鈥�dxdy$

Around the x- axis, the required region is symmetric.

That is,

$A=2{鈭�}_{x=1/2}^{x=1}鈥�{鈭�}_{y=0}^{y=\sqrt{1鈭�x^2}}鈥�dxdy$

Integrate first with respect to the y,$\mathrm{A}={鈭�}_{\mathrm{x}=1/2}^{\mathrm{x}=1}鈥�{鈭�}_{\mathrm{y}=鈭�\sqrt{1鈭�{\mathrm{x}}^2}}^{\mathrm{y}=\sqrt{1鈭�{\mathrm{x}}^2}}鈥�\mathrm{dxdy}$

$\begin{array}{r}\mathrm{A}=2{鈭�}_{\mathrm{x}=1/2}^{\mathrm{x}=1}鈥�[\mathrm{y}{]}_0^{\sqrt{1鈭�{\mathrm{x}}^2}}\mathrm{dx}\\ \mathrm{A}=2{鈭�}_{\mathrm{x}=1/2}^{\mathrm{x}=1}鈥�\sqrt{1鈭�{\mathrm{x}}^2}\mathrm{dx}\\ \mathrm{A}=2{\left[\frac{\mathrm{x}}{2}\sqrt{1鈭�{\mathrm{x}}^2}+\frac{1}{2}{\sin }^{鈭�1}鈦�\mathrm{x}\right]}_{1/2}^1\end{array}$

Set the boundaries

$\begin{array}{r}\mathrm{A}=2\left[\left\{\frac{1}{2}{\sin }^{鈭�1}鈦�1\right\}鈭�\left\{\frac{1/2}{2}\sqrt{1鈭�(1/2{)}^2}+\frac{1}{2}{\sin }^{鈭�1}鈦�(1/2)\right\}\right]\\ \mathrm{A}=2\left[\left\{\frac{1}{2},\frac{\mathrm{蟺}}{2}\right\}鈭�\left\{\frac{1}{4},\frac{\sqrt{3}}{2}+\frac{1}{2}鈰�\frac{\mathrm{蟺}}{6}\right\}\right]\\ \mathrm{A}=2\left[\frac{\mathrm{蟺}}{6}鈭�\frac{\sqrt{3}}{8}\right]\\ \mathrm{A}=\frac{\mathrm{蟺}}{3}鈭�\frac{\sqrt{3}}{4}\end{array}$

As a result, the needed region's area is$A=\frac{蟺}{3}鈭�\frac{\sqrt{3}}{4}$