check my ques,(warning math ahead)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

check my ques,(warning math ahead)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Question?
...
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 \[12\int\limits_{-a}^{a}[x^2[\frac{y}{2}\sqrt{t^2-y^2}+\frac{t^2}{2}\sin^{-1}\frac{y}{t}]_{0}^{t}+\int\limits_{0}^{\frac{\pi}{2}}t^2\sin^2(\theta).t^2\cos^2(\theta)d \theta+...\] \[...+\frac{1}{3}\int\limits_{0}^{\frac{\pi}{2}}t^3\cos^3(\theta).t\cos(\theta)d \theta]dx\] \[12 \int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{4}\int\limits_{0}^{\frac{\pi}{2}}\sin^2(2\theta)d \theta+\frac{t^4}{3}\int\limits_{0}^{\frac{\pi}{2}}\cos^4(\theta)d \theta]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta+\frac{t^4}{3}\int\limits_{0}^{\frac{\pi}{2}}(\cos^2(\theta)(1-\sin^2(\theta)))d \theta]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta+\frac{t^4}{3}\int\limits_{0}^{\frac{\pi}{2}}(\cos^2(\theta)-\cos^2(\theta)\sin^2(\theta)) d \theta]dx\]\[12 \int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta+\frac{t^4}{6}\int\limits_{0}^{\frac{\pi}{2}}(1+\cos(2\theta)-\frac{1}{2}\sin^2(2 \theta))d \theta]dx\] \[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta+\frac{t^4}{6}[\int\limits_{0}^{\frac{\pi}{2}}(1+\cos(2\theta))d \theta-\frac{1}{4}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta]]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^4}{4}+\frac{t^4}{6}\int\limits_{0}^{\frac{\pi}{2}}(1+\cos(2\theta))d \theta+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta-\frac{t^4}{24}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{\pi t^4}{12}+\frac{t^4}{8}\int\limits_{0}^{\frac{\pi}{2}}((1-\cos(4\theta))d \theta)(1-\frac{1}{3})]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{\pi t^4}{12}+\frac{t^4}{8}.\frac{2}{3}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta]dx\]\[12\int\limits_{-a}^{a}[\frac{\pi x^2t^2}{4}+\frac{\pi t^4}{12}+\frac{\pi t^4}{24}]dx=3\pi \int\limits_{-a}^{a}(x^2t^2+\frac{t^4}{3}+\frac{t^4}{6})dx\]\[3\pi \int\limits_{-a}^{a}[x^2(a^2-x^2)+\frac{(a^2-x^2)^2}{2}]dx=6\pi [\int\limits_{0}^{a}(a^2x^2-x^4)dx+\frac{1}{2}\int\limits_{0}^{a}(a^4-2a^2x^2+x^4)dx]\]\[6\pi[\frac{a^5}{3}-\frac{a^5}{5}+\frac{1}{2}(a^5-\frac{2}{3}a^5+\frac{a^5}{5})]\]\[6\pi[\frac{a^5}{3}-\frac{a^5}{5}+\frac{a^5}{2}-\frac{a^5}{3}+\frac{a^5}{10}]=6\pi[\frac{-2a^5+5a^5+a^5}{10}]=6\pi[\frac{4a^5}{10}]=\frac{24\pi a^5}{10}=\frac{12\pi a^5}{5}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

oh no its cut by the page
im getting \[\frac{12\pi a^5}{5}\] after a looong way
why haven't you switched to spherical coordinates ?
I haven't studied them yet :P
Im not doing this. **** this.
How fast is this with spherical?
cartesian to spherical : \(dx~dy~dz = \rho^2\sin\phi~ d\rho~ d\theta ~d\phi\)
\[\iiint_\limits V (3x^2+3y^2+3z^2)dV \\~\\ =\int\limits_{0}^{\pi}~\int\limits_{0}^{2\pi}~\int\limits_0^a (3\rho^2)~ \rho^2\sin\phi~ d\rho~ d\theta ~d\phi \\~\\ \] one or two more steps and you're done !
\[3 \iiint_\limits{V}\rho^2(\rho^2\sin(\theta)d \rho d \theta d \phi)=3\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int\limits_{-a}^{a}\rho^4 \sin(\theta) d \rho d \theta d \phi\]\[24\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{a}\rho^4 \sin(\theta) d \rho d \theta d \phi=\frac{24a^5}{5}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{\pi}{2}}\sin(\theta) d \theta d \phi\]\[\frac{24a^5}{5}\int\limits_{0}^{\frac{\pi}{2}}[-\cos(\theta)]_{0}^{\frac{\pi}{2}}d \phi=\frac{24a^5}{5}\int\limits_{0}^{\frac{\pi}{2}}1.d \phi=\frac{24a^5}{5}.\frac{\pi}{2}=\frac{12\pi a^5}{5}\]
is this good
that problems long without converting coordinates
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
sweet

Not the answer you are looking for?

Search for more explanations.

Ask your own question