Kripya ise solve kre!

1.	Kripya ise solve kre!
Answer» trong>S O L U T I O N : ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Let US take, ⠀⠀⠀⠀⠀⠀ x = aX, y = bY, z = cZ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Then, ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{\dfrac{\partial(x,y,z)}{\partial(X,Y,Z)}}=\left\|\begin{array}{ccc}\sf{a}&\sf{<klux>0</klux>}&\sf{0}\\ \sf{0}&\sf{b}&\sf{0}\\ \sf{0}&\sf{0}&\sf{c}\end{array}\right\|\sf{=abc\neq 0}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\therefore\sf{\int\int\int_{V}\sqrt{\Big[1-\Big(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}\Big)\Big]}\:dx\:dy\:dz}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=abc\int\int\int_{V'}\sqrt{\Big[1-\Big(\dfrac{(aX)^{2}}{a^{2}}+\dfrac{(bY)^{2}}{b^{2}}+\dfrac{(cZ)^{2}}{c^{2}}\Big)\Big]}\:dX\:dY\:dZ}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=abc\int\int\int_{V'}\sqrt{[1-(X^{2}+Y^{2}+Z^{2})]}\:dX\:dY\:dZ}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Where V' is the region bounded by X² + Y² + Z² = 1. ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Again we take, ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{\;\;\;\;\;\bullet\;\;\;X=r\:sin\theta\:cos\phi}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{\;\;\;\;\;\bullet\;\;\;Y=r\:sin\theta\:sin\phi}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{\;\;\;\;\;\bullet\;\;\;Z=r\:cos\theta}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Then, ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{\dfrac{\partial(X,Y,Z)}{\partial(r,\theta,\phi)}=r^{2}\:sin\theta>0}$ except when X = Y = Z = 0 ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ The new region be V" = [0, 1; 0, π; 0, 2π] ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Now, X² + Y² + Z², ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=r^{2}\:sin^{2}\theta\:cos^{2}\phi+r^{2}\:sin^{2}\theta\:sin^{2}\phi+r^{2}\:cos^{2}\theta}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=r^{2}\:sin^{2}\theta\:(cos^{2}\phi+sin^{2}\phi)+r^{2}\:cos^{2}\theta}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=r^{2}\:(sin^{2}\theta+cos^{2}\theta)}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=r^{2}}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Continuing the integration, we get, ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=abc\int\int\int_{V''}\sqrt{1-r^{2}}\:r^{2}\:sin\theta\:<klux>DR</klux>\:d\theta\:d\phi}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=abc\int_{0}^{1}r^{2}\sqrt{1-r^{2}}\:dr\int_{0}^{\pi}sin\theta\:d\theta\int_{0}^{2\pi}d\phi}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=abc\int_{0}^{1}r^{2}\sqrt{1-r^{2}}\:dr\:\times(2\times 2\pi)}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=4\pi\:abc\int_{0}^{1}r^{2}\sqrt{1-r^{2}}\:dr}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ [Put r = sint, then dr = cost dt. When r changes from 0 to 1, t changes from 0 to π/2] ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=4\pi\:abc\int_{0}^{\large\frac{\pi}{2}}sin^{2}t\:cos^{2}t\:dt}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=\dfrac{1}{2}\pi\:abc\int_{0}^{\large\frac{\pi}{2}}2\:sin^{2}2t\:dt}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=\dfrac{1}{2}\pi\:abc\int_{0}^{\large\frac{\pi}{2}}(1-cos4t)\:dt}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=\dfrac{1}{2}\pi\:abc\:\bigg[({\dfrac{\pi}{2}}-1)-(0-1)\bigg]}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\sf{=\dfrac{1}{4}{\pi}^{2}\:abc}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\displaystyle\therefore\;\sf{\int\int\int_{V}\sqrt{\Big[1-\Big(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}\Big)\Big]}\:dx\:dy\:dz=\dfrac{{\pi}^{2}\:abc}{4}}$

Answer»

trong>S O L U T I O N :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Let US take,

⠀⠀⠀⠀⠀⠀

x = aX, y = bY, z = cZ

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Then,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{\dfrac{\partial(x,y,z)}{\partial(X,Y,Z)}}=\left|\begin{array}{ccc}\sf{a}&\sf{<klux>0</klux>}&\sf{0}\\ \sf{0}&\sf{b}&\sf{0}\\ \sf{0}&\sf{0}&\sf{c}\end{array}\right|\sf{=abc\neq 0}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\therefore\sf{\int\int\int_{V}\sqrt{\Big[1-\Big(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}\Big)\Big]}\:dx\:dy\:dz}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=abc\int\int\int_{V'}\sqrt{\Big[1-\Big(\dfrac{(aX)^{2}}{a^{2}}+\dfrac{(bY)^{2}}{b^{2}}+\dfrac{(cZ)^{2}}{c^{2}}\Big)\Big]}\:dX\:dY\:dZ}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=abc\int\int\int_{V'}\sqrt{[1-(X^{2}+Y^{2}+Z^{2})]}\:dX\:dY\:dZ}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Where V' is the region bounded by X² + Y² + Z² = 1.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Again we take,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{\;\;\;\;\;\bullet\;\;\;X=r\:sin\theta\:cos\phi}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{\;\;\;\;\;\bullet\;\;\;Y=r\:sin\theta\:sin\phi}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{\;\;\;\;\;\bullet\;\;\;Z=r\:cos\theta}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Then,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{\dfrac{\partial(X,Y,Z)}{\partial(r,\theta,\phi)}=r^{2}\:sin\theta>0}$ except when X = Y = Z = 0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

The new region be V" = [0, 1; 0, π; 0, 2π]

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Now, X² + Y² + Z²,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=r^{2}\:sin^{2}\theta\:cos^{2}\phi+r^{2}\:sin^{2}\theta\:sin^{2}\phi+r^{2}\:cos^{2}\theta}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=r^{2}\:sin^{2}\theta\:(cos^{2}\phi+sin^{2}\phi)+r^{2}\:cos^{2}\theta}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=r^{2}\:(sin^{2}\theta+cos^{2}\theta)}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=r^{2}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Continuing the integration, we get,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=abc\int\int\int_{V''}\sqrt{1-r^{2}}\:r^{2}\:sin\theta\:<klux>DR</klux>\:d\theta\:d\phi}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=abc\int_{0}^{1}r^{2}\sqrt{1-r^{2}}\:dr\int_{0}^{\pi}sin\theta\:d\theta\int_{0}^{2\pi}d\phi}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=abc\int_{0}^{1}r^{2}\sqrt{1-r^{2}}\:dr\:\times(2\times 2\pi)}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=4\pi\:abc\int_{0}^{1}r^{2}\sqrt{1-r^{2}}\:dr}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

[Put r = sint, then dr = cost dt. When r changes from 0 to 1, t changes from 0 to π/2]

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=4\pi\:abc\int_{0}^{\large\frac{\pi}{2}}sin^{2}t\:cos^{2}t\:dt}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=\dfrac{1}{2}\pi\:abc\int_{0}^{\large\frac{\pi}{2}}2\:sin^{2}2t\:dt}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=\dfrac{1}{2}\pi\:abc\int_{0}^{\large\frac{\pi}{2}}(1-cos4t)\:dt}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=\dfrac{1}{2}\pi\:abc\:\bigg[({\dfrac{\pi}{2}}-1)-(0-1)\bigg]}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\sf{=\dfrac{1}{4}{\pi}^{2}\:abc}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle\therefore\;\sf{\int\int\int_{V}\sqrt{\Big[1-\Big(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}\Big)\Big]}\:dx\:dy\:dz=\dfrac{{\pi}^{2}\:abc}{4}}$

Kripya ise solve kre!

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Kripya ise solve kre!