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$$\int\limits_0^5 \int\limits_{\sqrt{25-x^2}}^{-\sqrt{25-x^2}} \int\limits_{\sqrt{25-x^2-z^2}}^{-\sqrt{25-x^2-z^2}} \frac{1}{x^2+y^2+z^2} \,\mathrm dy\,\mathrm dz\,\mathrm dx$$

triple integral trying to change to spherical coordinates.

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    What is your question exactly? Where are you stuck?2012-12-02
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    so i have to change it to spherical coordinates2012-12-02
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    and i tried doing it and got 50*pi. but the correct answer is 25* pi2012-12-02
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    theta is from 0 to 2pi. rho from 0 to 5. phi from 0 to pi. these are limits i chose. and the function is 1/(rho^2)2012-12-02
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    You should type (in [LaTeX](http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117)) all your steps out, then someone can check them. It's impossible to see where your error is without seeing your work.2012-12-02
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    sorry i am new. i will try that. thank you2012-12-02
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    Note that $ x $ doesn't vary from -5 to 5, it varies from 0 to 5. Your limits in spherical are thus incorrect, and this is why you are off by a factor of 2.2012-12-02
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    @AAbc: Are you sure that you copied the question correctly. I think you are missing the square root. The integrand should be $\frac{1}{\sqrt{x^2+y^2+z^2}}$.2012-12-02

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The integrals on $y$ and $z$ have their limits in an unusual way (positive below, negative above) but changing both at the same time won't change the value of the integral (as each contributes a minus sign). So we want $$ I=\int_0^5 \int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}} \int_{-\sqrt{25-x^2-z^2}}^{\sqrt{25-x^2-z^2}} \frac{1}{x^2+y^2+z^2} \,dy~dz~dx $$ The region is the half solid sphere of radius $5$ centered at the origin, with $x\geq0$. In spherical coordinates, this is $$ 0\leq\rho\leq5,\ -\pi/2\leq\theta\leq\pi/2,\ 0\leq\phi\leq\pi. $$ So $$ I=\int_{\pi/2}^{\pi/2}\int_0^{\pi}\int_0^5\frac1{\rho^2}\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta=\int_{\pi/2}^{\pi/2}\int_0^{\pi}\int_0^5\sin\phi\,d\rho\,d\phi\,d\theta=5\pi\,\int_0^\pi\sin\phi\,d\phi=5\pi\,(-\cos\phi)|_0^\pi=10\pi. $$

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    The equation in the integrand should be $\frac{1}{x^2+y^2+z^2} = \frac{1}{{\rho}^2}$.2012-12-02
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    You right. But as you said in your other comment, most likely the square root is missing, then. I've edited so far without the square root.2012-12-02
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    I got exactly the same answer as you, but I did not post it.2012-12-02