Question

solve this question. ( triple integrals) ​

Answer 1

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

Step-by-step explanation:

Let us take

x = aX, y = bY, z = cZ

Then,

$\displaystyle\mathrm{\frac{\partial(x,y,z)}{\partial(X,Y,Z)}}=\left|\begin{array}{ccc}\mathrm{a}&\mathrm{0}&\mathrm{0}\\ \mathrm{0}&\mathrm{b}&\mathrm{0}\\ \mathrm{0}&\mathrm{0}&\mathrm{c}\end{array}\right|\mathrm{=abc\neq 0}$

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

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

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

where V' is the region bounded by $\displaystyle\mathrm{X^{2}+Y^{2}+Z^{2}=1}$

Again we take

$\displaystyle\mathrm{X=r\:sin\theta\:cos\phi}$

$\displaystyle\mathrm{Y=r\:sin\theta\:sin\phi}$

$\displaystyle\mathrm{Z=r\:cos\theta}$

Then,

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

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

Now, $\displaystyle\mathrm{X^{2}+Y^{2}+Z^{2}}$

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

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

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

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

Continuing the integration, we get

$\displaystyle\mathrm{abc\int\int\int_{V''}\sqrt{1-r^{2}}\:r^{2}\:sin\theta\:dr\:d\theta\:d\phi}$

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

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

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

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

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

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

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

$\displaystyle\mathrm{=\frac{1}{2}\pi\:abc\:[(\frac{\pi}{2}-1)-(0-1)]}$

$\displaystyle\mathrm{=\frac{1}{4}{\pi}^{2}\:abc}$

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

Answer 2

Step-by-step explanation:

Hope this attachment may help u.....