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 April 28th, 2017, 04:51 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Triple Integral of a sphere question Compute ∫ ∫ ∫ x + y + z dV over the region x^2 + y^2 + z^2 <= 1 in the First octant. Answer: 3pi/16 I would be much appreciated if anyone show me how to define the boundaries of this integral. Thanks. Last edited by zollen; April 28th, 2017 at 05:22 PM. April 28th, 2017, 06:17 PM #2 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 Depends on where you are at in your class, doing it in Cartesian can be complicated and tedious. If you have gone through spherical coordinates, that that would work best, imo. $\displaystyle r(x,y,z)=r(\rho,\phi,\theta)=(\rho\sin\phi\cos \theta,\rho \sin\phi\sin\theta,\rho\cos\phi)$ Radius of the sphere $\displaystyle 0\leq\rho\leq 1$ First Octet $\displaystyle 0\leq\phi\leq \frac{\pi}{2}$ $\displaystyle 0\leq\theta\leq \frac{\pi}{2}$ $\displaystyle \int_{0}^{ \frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{1} f(\rho,\phi,\theta)\rho^2\sin\phi d\rho d\phi d\theta$ $\displaystyle \int_{0}^{ \frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{1} \rho^3(\sin^2\phi\cos\theta+\sin^2\phi\sin\theta+\ sin\phi\cos\phi)d\rho d\phi d\theta$ April 28th, 2017, 06:18 PM   #3
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Thanks for the suggestion. But I have not learned spherical or cylindrical coordinates yet. I want to solve this with only Cartesian coordinate.

Quote:
 Originally Posted by phrack999 Depends on where you are at in your class, doing it in Cartesian can be complicated and tedious. If you have gone through spherical coordinates, that that would work best, imo. $\displaystyle r(x,y,z)=r(\rho,\phi,\theta)=(\rho\sin\phi\cos \theta,\rho \sin\phi\sin\theta,\rho\cos\phi)$ Radius of the sphere $\displaystyle 0\leq\rho\leq 1$ First Octet $\displaystyle 0\leq\phi\leq \frac{\pi}{2}$ $\displaystyle 0\leq\theta\leq \frac{\pi}{2}$ $\displaystyle \int_{0}^{ \frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{1} f(\rho,\phi,\theta)\rho^2\sin\phi d\rho d\phi d\theta$ $\displaystyle \int_{0}^{ \frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{1} \rho^3(\sin^2\phi\cos\theta+\sin^2\phi\sin\theta+\ sin\phi\cos\phi)d\rho d\phi d\theta$ April 28th, 2017, 06:34 PM #4 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 Well then I supposed you'd have to go: $\displaystyle 0\leq z\leq \sqrt{1-x^2-y^2}$ $\displaystyle 0\leq y\leq \sqrt{1-x^2}$ $\displaystyle 0\leq x\leq 1$ $\displaystyle \int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\int_{0}^{\sqrt{1-x^2-y^2}}(x+y+z)dzdydx$ And that's a whole world of hurt dealing with those integrals. The trick I try to remember is $\displaystyle \sqrt{\bigg(\sqrt{1-x^2}\bigg)^2-y^2}$ for trigonometric substitution or $\displaystyle \sqrt{a^2-x^2}$ if you have a table of integrals. Still, going to be unpleasant. Hopefully, someone else will have a better method, but that is why they made spherical coordinates, to simplify this. Last edited by phrack999; April 28th, 2017 at 06:59 PM. April 28th, 2017, 07:28 PM #5 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 May I ask how did you come up the upper limit of y and z? The upper limit of y seems to be missing the z component (i.e. sqrt(1-x^2-z^2) ) But it would not make sense to place z there because z has already been part of the first integral. That's why I was stuck... April 28th, 2017, 08:35 PM #6 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 $\displaystyle z=\sqrt{1-x^2-y^2}$ Thats why its the inner integral, because your accounting for the z at that point. With y, you project it onto the xy-plane by setting z=0: $\displaystyle 0=1-x^2-y^2$ $\displaystyle y=\sqrt{1-x^2}$ Then you need the range for x, which is simply the radius for the sphere: $\displaystyle 0\leq x\leq 1$ The zero lower limits come from the fact that it is in the first octet, so zero to the positive values of x,y,z. This was one of the harder parts of multi-variable calculus for me to wrap my brain around. Try to think of it as cumulative. With the inner integral, it is representing itself in terms of x,y, so that when the integration happens in x and y, it'll be taken in to account. So its encoding itself as a function of x and y into the the later integrals. Same holds true for y in this instance, its encoding itself in terms of x, so that when the integral of x is done, its in there as well as the z function. When dealing with volume integral of shapes don't think about the function that is going to be inside the integral, think of just how you would calculate the volume of that shape, $\displaystyle \int_{a}^{b}\int_{u(x)}^{v(x)}\int_{f(x,y)}^{g(x,y )} dV$ then once you have your ranges figured out, deal with the function be integrated across. $\displaystyle \int_{a}^{b}\int_{u(x)}^{v(x)}\int_{f(x,y)}^{g(x,y )} f(x,y,z)dV$ Thanks from zollen Last edited by phrack999; April 28th, 2017 at 08:46 PM. Tags integral, question, sphere, triple Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post pebblepigs Calculus 3 April 2nd, 2015 09:10 PM laura123 Calculus 5 May 9th, 2013 10:11 PM triplekite Calculus 1 December 5th, 2012 08:16 AM tinyone Calculus 0 November 7th, 2010 08:54 AM sp1227 Calculus 0 March 22nd, 2008 11:31 AM

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