91影视

The objective of this problem is to construct an example of area of a circle with radius $R$.

Find the area of a lawn grazed out by a cow. The cow is tied with a rope of length $R$meter. Consider a circle of radius $R$

Area of a sector of a circle of radius $R$enclosed by $胃=伪$ and $胃=尾$can be expressed as $\frac{1}{2}{鈭�}_{胃-伪}^{胃-尾}{r}^{2}d胃$
Area of a quarter of a circle can be expressed as a sum of area of four sectors.
Area of a circle $=$ Sum of area of four quarters
Area of a quarter of a circle $=\frac{1}{2}{鈭�}_0^{蟺/2}{r}^{2}d胃$
Therefore, area of a circular ground
$A=4脳\frac{1}{2}{鈭�}_0^{蟺/2}{r}^{2}d胃$

$A=4脳\frac{1}{2}{鈭�}_0^{蟺/2}{R}^{2}d胃[r=R]$

$A=2R^2{鈭�}_0^{蟺/2}d胃$

Integrate with respect to $胃$.
$A=2R^2[胃{]}_0^{*/2}$
Put the limits
$A=2R^2\left[\frac{蟺}{2}-0\right]$

$A=蟺R^2$
Thus, the area of a circular ground is
$A=蟺R^2$

Take an example of sphere of radius } a \text { and center at origin.

The equation of sphere is
$x^2+y^2+z^2=a^2$
The sphere is symmetrical about $x-y$ plane. Therefore, its volume can be computed as the upper half of sphere multiplied by 2
$V=2{鈭�}_{x-a}^{x-a}{鈭�}_{y-0}^{\sqrt{a^2-x^2}}zdxdy$

$V=2{鈭�}_{x-a}^{x-a}{鈭�}_{y=0}^{\sqrt{a^2-x^2}}\sqrt{a^2-\left(x^2+y^2\right)}dxdy\left[x^2+y^2+z^2=a^2\right]$
For the polar form
Substitute $x=r\cos 胃,y=r\sin 胃$and $dxdy=rdrd胃$
Where, $0鈮�r鈮�a$and $-蟺鈮�胃鈮�蟺$
$V=2{鈭�}_{胃=-蟺}^{胃-s}{鈭�}_{r=0}^{r-a}\sqrt{a^2-\left(r^2{\cos }^2胃+r^2{\sin }^2胃\right)}rdrd胃$

$V=2{鈭�}_{x=-a}^{x-a}{鈭�}_{y=0}^{\sqrt{a^2-x^2}}\sqrt{a^2-\left(x^2+y^2\right)}dxdy\left[x^2+y^2+z^2=a^2\right]$
$V=2{鈭�}_{胃-n}^{胃-蟺}{鈭�}_{r=0}^{r=a}\sqrt{a^2-r^2}rdrd胃\left[{\sin }^2胃+{\cos }^2胃=1\right]$
Integrate with respect to $r$.

$V=2{鈭�}_{胃-蟺}^{胃-蟺}{\left[-\frac{1}{3}{\left(a^2-r^2\right)}^{3/2}\right]}_0^{a}d胃$
$V=2{鈭�}_{胃-蟺}^{胃-蟺}\left[\frac{1}{3}{\left(a^2\right)}^{3/2}\right]d胃$

$V=\frac{2}{3}{a}^3{鈭�}_{胃-蟺}^{胃-蟺}d胃$

Integrate with respect to $胃$.
$V=\frac{2}{3}{a}^3[胃{]}_{-z}^z$

$V=\frac{2}{3}{a}^3[蟺-(-蟺\left)\right]$

$V=\frac{4}{3}蟺a^3$

Thus, the volume of a sphere is
$V=\frac{4}{3}蟺a^3$

Consider an example of double integration.
$I={鈭�}_0^2{鈭�}_0^{\sqrt{2x-x^2}}\left(x^2+y^2\right)dydx$

$\backslash begin\left\{aligned\right\}\&I=\backslash int\_\left\{0\right\}^\{\backslash pi/2\}\backslash int\_\left\{0\right\}^\{2\backslash \cos \backslash theta\}\backslash left(r^\{2\}\backslash \cos ^\{2\}\backslash theta+r^\{2\}\backslash \sin ^\{2\}\backslash theta\backslash right)rdrd\backslash theta\backslash \backslash \&I=\backslash int\_\left\{0\right\}^\{x/2\}\backslash int\_\left\{0\right\}^\{2\backslash \cos \backslash theta\}r^\left\{3\right\}drd\backslash theta\backslash \backslash \&I=\backslash int\_\left\{0\right\}^\{\backslash pi/2\}\backslash left[\backslash frac\{r^\left\{4\right\}\left\}\right\{4\}\backslash right]\_\left\{0\right\}^\{2\backslash operatorname\{ees\}\backslash theta\}d\backslash theta\backslash \backslash \&I=\backslash int\_\left\{0\right\}^\{\backslash pi/2\}\backslash left[\backslash frac\{16\left\}\right\{4\}\backslash \cos ^\{4\}\backslash theta-0\backslash right]d\backslash theta\backslash \backslash \&I=4\backslash int\_\left\{0\right\}^\{\backslash pi/2\}\backslash \cos ^\left\{4\right\}\backslash thetad\backslash theta\backslash \backslash \&I=4\backslash int\_\left\{0\right\}^\{\backslash pi/2\}\backslash left(\backslash \cos ^\{2\}\backslash theta\backslash right)^\left\{2\right\}d\backslash theta\backslash \backslash \&I=4\backslash int\_\left\{0\right\}^\{\backslash pi/2\}\backslash left(\backslash frac\{1+\backslash \cos 2\backslash theta\left\}\right\{2\}\backslash right)^\left\{2\right\}d\backslash theta\backslash \backslash \&I=4\backslash int\_\left\{0\right\}^\{\backslash pi/2\}\backslash left(\backslash frac\{1+2\backslash \cos 2\backslash theta+\backslash \cos ^\left\{2\right\}2\backslash theta\left\}\right\{4\}\backslash right)d\backslash theta\backslash \backslash \&I=\backslash int\_\left\{0\right\}^\{\backslash pi/2\}\backslash left\backslash \{1+2\backslash \cos 2\backslash theta+\backslash frac\{1\left\}\right\{2\left\}\right(1+\backslash \cos 4\backslash theta)\backslash right\backslash \}d\backslash theta\backslash \backslash \&I=\backslash left[\backslash theta+\backslash \sin 2\backslash theta+\backslash frac\{1\left\}\right\{2\}\backslash left(\backslash theta+\backslash frac\left\{1\right\}\left\{4\right\}\backslash \sin 4\backslash theta\backslash right)\backslash right]\_\left\{0\right\}^\{\backslash pi/2\}\backslash \backslash \&I=\backslash left[\backslash frac\{\backslash pi\left\}\right\{2\}+\backslash \sin (\backslash pi)+\backslash frac\{1\left\}\right\{2\}\backslash left(\backslash frac\{\backslash pi\}\left\{2\right\}+\backslash frac\left\{1\right\}\left\{4\right\}\backslash \sin 2\backslash pi\backslash right)\backslash right]\backslash end\left\{aligned\right\}$

Here upper limit of $y$is

$y=\sqrt{2x-x^2}$

$y^2=2x-x^2\text{[Square both sides]}$
$鈬�x^2+y^2+2x=0鈥�鈥�\text{(1)}$

Equation (1) represents a circle with center (1,0) and unit radius.

Lower limit of $y$ is 0 .
Region of integration is the upper half of circle.
Substitute $x=r\cos 胃,y=r\sin 胃$ in equation (1)
$r^2-2r\cos 胃=0鈬�r(r-2\cos 胃)=0$

$r=0,r=2\cos 胃$

$胃=0,胃=\frac{蟺}{2}$
Therefore,

$I={鈭�}_0^{蟺/2}{鈭�}_0^{2\cos 胃}\left(r^2{\cos }^2胃+r^2{\sin }^2胃\right)rdrd胃$
$I={鈭�}_0^{蟺/2}{鈭�}_0^{2\cos 胃}{r}^{3}drd胃$
$I={鈭�}_0^{蟺/2}{\left[\frac{r^4}{4}\right]}_0^{2\cos 胃}d胃$
$I={鈭�}_0^{蟺/2}\left[\frac{16}{4}{\cos }^4胃-0\right]d胃$

$I=4{鈭�}_0^{蟺/2}{\cos }^4胃d胃$

$I=4{鈭�}_0^{蟺/2}{\left({\cos }^2胃\right)}^{2}d胃$

$I=4{鈭�}_0^{蟺/2}{\left(\frac{1+\cos 2胃}{2}\right)}^{2}d胃$

$I={\left[{鈭�}_0^{蟺/2}\left(\sin \left(蟺\right)+\frac{1}{2}\left(\frac{蟺}{2}+\frac{1}{4}\sin 2蟺\right)\right]\sin 2胃+\frac{1}{2}\left(胃+\frac{1}{4}\sin 4胃\right)\right]}_0^{蟺/2}$

$I=\left[\frac{1+2\cos 2胃+{\cos }^{2}2胃}{4}d胃\right.$

$I=\left[\frac{蟺}{2}+\sin \left(蟺\right)+\frac{1}{2}\left(\frac{蟺}{2}+\frac{1}{4}\sin 2蟺\right)\right]$

$I=\frac{3蟺}{4}$

${鈭�}_0^2{鈭�}_0^{\sqrt{2x-x^2}}\left(x^2+y^2\right)dydx=\frac{3蟺}{4}$

True