Question

check my ques,(warning math ahead)

Answer 1

Question?

Answer 2

...

Answer 3

Use divergence theorem to evaluate
\[\iint_\limits S \vec A.d \vec S \space \space \space ; \space \space \space \vec A=x^3\hat i+y^3 \hat j+z^3 \hat k\]
S is the surface of the sphere
\[x^2+y^2+z^2=a^2\]
From divergence theorem we have,
\[\iint_\limits S \vec A .d \vec S=\iiint_\limits V (\vec \nabla . \vec A)dV=\iiint_\limits V (3x^2+3y^2+3z^2)dV\]\[3 \int\limits_{-a}^{a}[\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}[\int\limits_{-\sqrt{a^2-(x^2+y^2)}}^{\sqrt{a^2-(x^2+y^2)}}(x^2+y^2+z^2)dz]dy]dx\]
\[3 \int\limits_{-a}^{a}[\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}[2(x^2+y^2)\int\limits_{0}^{\sqrt{a^2-(x^2+y^2)}}dz+2\int\limits_{0}^{\sqrt{a^2-(x^2+y^2)}}z^2dz]dy]dx\]\[6\int\limits_{-a}^{a}[\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}[(x^2+y^2)\sqrt{a^2-(x^2+y^2)}+\frac{(a^2-(x^2+y^2))^\frac{3}{2}}{3}]dy]dx\]
\[6\int\limits_{-a}^{a}[2x^2\int\limits_{0}^{\sqrt{a^2-x^2}}\sqrt{a^2-(x^2+y^2)}dy+2\int\limits_{0}^{\sqrt{a^2-x^2}}y^2\sqrt{a^2-(x^2+y^2)}...\]\[...+\frac{2}{3}\int\limits_{0}^{\sqrt{a^2-x^2}}(a^2-(x^2+y^2))^\frac{3}{2}dy]dx\]
Let
\[a^2-x^2=t^2\]
\[12\int\limits_{-a}^{a}[x^2\int\limits_{0}^{t}\sqrt{t^2-y^2}dy+\int\limits_{0}^{t}y^2\sqrt{t^2-y^2}dy+\frac{1}{3}\int\limits_{0}^{t}(t^2-y^2)^\frac{3}{2}dy]dx\]
Using the substitution
\[y=t\sin(\theta)\]
whereever neccessary
\[12\int\limits_{-a}^{a}[x^2[\frac{y}{2}\sqrt{t^2-y^2}+\frac{t^2}{2}\sin^{-1}\frac{y}{t}]_{0}^{t}+\int\limits_{0}^{\frac{\pi}{2}}t^2\sin^2(\theta).t^2\cos^2(\theta)d \theta+...\]
\[...+\frac{1}{3}\int\limits_{0}^{\frac{\pi}{2}}t^3\cos^3(\theta).t\cos(\theta)d \theta]dx\]
\[12 \int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{4}\int\limits_{0}^{\frac{\pi}{2}}\sin^2(2\theta)d \theta+\frac{t^4}{3}\int\limits_{0}^{\frac{\pi}{2}}\cos^4(\theta)d \theta]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta+\frac{t^4}{3}\int\limits_{0}^{\frac{\pi}{2}}(\cos^2(\theta)(1-\sin^2(\theta)))d \theta]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta+\frac{t^4}{3}\int\limits_{0}^{\frac{\pi}{2}}(\cos^2(\theta)-\cos^2(\theta)\sin^2(\theta)) d \theta]dx\]\[12 \int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta+\frac{t^4}{6}\int\limits_{0}^{\frac{\pi}{2}}(1+\cos(2\theta)-\frac{1}{2}\sin^2(2 \theta))d \theta]dx\]
\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta+\frac{t^4}{6}[\int\limits_{0}^{\frac{\pi}{2}}(1+\cos(2\theta))d \theta-\frac{1}{4}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta]]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^4}{4}+\frac{t^4}{6}\int\limits_{0}^{\frac{\pi}{2}}(1+\cos(2\theta))d \theta+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta-\frac{t^4}{24}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{\pi t^4}{12}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}((1-\cos(4\theta))d \theta)(1-\frac{1}{3})]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{\pi t^4}{12}+\frac{t^4}{8}.\frac{2}{3}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{\pi t^4}{12}+\frac{\pi t^4}{24}]dx=3\pi \int\limits_{-a}^{a}(x^2t^2+\frac{t^4}{3}+\frac{t^4}{6})dx\]\[3\pi \int\limits_{-a}^{a}[x^2(a^2-x^2)+\frac{(a^2-x^2)^2}{2}]dx=6\pi [\int\limits_{0}^{a}(a^2x^2-x^4)dx+\frac{1}{2}\int\limits_{0}^{a}(a^4-2a^2x^2+x^4)dx]\]\[6\pi[\frac{a^5}{3}-\frac{a^5}{5}+\frac{1}{2}(a^5-\frac{2}{3}a^5+\frac{a^5}{5})]\]\[6\pi[\frac{a^5}{3}-\frac{a^5}{5}+\frac{a^5}{2}-\frac{a^5}{3}+\frac{a^5}{10}]=6\pi[\frac{-2a^5+5a^5+a^5}{10}]=6\pi[\frac{4a^5}{10}]=\frac{24\pi a^5}{10}=\frac{12\pi a^5}{5}\]

Answer 4

oh no its cut by the page

Answer 5

im getting \[\frac{12\pi a^5}{5}\]
after a looong way

Answer 6

why haven't you switched to spherical coordinates ?

Answer 7

I haven't studied them yet :P

Answer 8

Im not doing this. **** this.

Answer 9

How fast is this with spherical?

Answer 10

cartesian to spherical :
\(dx~dy~dz = \rho^2\sin\phi~ d\rho~ d\theta ~d\phi\)

Answer 11

\[\iiint_\limits V (3x^2+3y^2+3z^2)dV \\~\\
=\int\limits_{0}^{\pi}~\int\limits_{0}^{2\pi}~\int\limits_0^a (3\rho^2)~ \rho^2\sin\phi~ d\rho~ d\theta ~d\phi \\~\\
\]

one or two more steps and you're done !

Answer 12

\[3 \iiint_\limits{V}\rho^2(\rho^2\sin(\theta)d \rho d \theta d \phi)=3\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int\limits_{-a}^{a}\rho^4 \sin(\theta) d \rho d \theta d \phi\]\[24\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{a}\rho^4 \sin(\theta) d \rho d \theta d \phi=\frac{24a^5}{5}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{\pi}{2}}\sin(\theta) d \theta d \phi\]\[\frac{24a^5}{5}\int\limits_{0}^{\frac{\pi}{2}}[-\cos(\theta)]_{0}^{\frac{\pi}{2}}d \phi=\frac{24a^5}{5}\int\limits_{0}^{\frac{\pi}{2}}1.d \phi=\frac{24a^5}{5}.\frac{\pi}{2}=\frac{12\pi a^5}{5}\]

Answer 13

is this good

Answer 14

that problems long without converting coordinates

Answer 15

you may double check with wolfram widget
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=89c969c21b169fa996f899d9b2a98588&title=Spherical%20Integral%20Calculator&theme=blue&i0=3rho%5E4sin(phi)&i1=drho%20dtheta%20dphi&i2=0&i3=a&i4=0&i5=pi&i6=0&i7=2pi&podSelect=&showAssumptions=1&showWarnings=1

Answer 16

sweet