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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? –  icurays1 Dec 2 '12 at 0:03
    
so i have to change it to spherical coordinates –  AAbc Dec 2 '12 at 0:22
    
and i tried doing it and got 50*pi. but the correct answer is 25* pi –  AAbc Dec 2 '12 at 0:23
    
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) –  AAbc Dec 2 '12 at 0:24
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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. –  R.R. Dec 2 '12 at 1:07

1 Answer 1

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 $\frac{1}{x^2+y^2+z^2} = \frac{1}{{\rho}^2}$. –  Mhenni Benghorbal Dec 2 '12 at 2:35
    
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. –  Martin Argerami Dec 2 '12 at 2:43
    
I got exactly the same answer as you, but I did not post it. –  Mhenni Benghorbal Dec 2 '12 at 3:05

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