## TomLikesPhysics Group Title I need help with an integral. one year ago one year ago

1. ajit.shirsat Group Title

which

2. amistre64 Group Title

$\begin{array}l\color{red}{\text{I}}\color{orange}{\text{n}}\color{#e6e600}{\text{t}}\color{green}{\text{e}}\color{blue}{\text{g}}\color{purple}{\text{r}}\color{purple}{\text{a}}\color{red}{\text{t}}\color{orange}{\text{i}}\color{#e6e600}{\text{o}}\color{green}{\text{n}}\color{blue}{\text{ }}\color{purple}{\text{n}}\color{purple}{\text{e}}\color{red}{\text{e}}\color{orange}{\text{d}}\color{#e6e600}{\text{s}}\color{green}{\text{ }}\color{blue}{\text{n}}\color{purple}{\text{o}}\color{purple}{\text{ }}\color{red}{\text{h}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{p}}\color{blue}{\text{ }}\color{purple}{\text{.}}\color{purple}{\text{.}}\color{red}{\text{.}}\color{orange}{\text{.}}\color{#e6e600}{\text{ }}\color{green}{\text{✌}}\color{blue}{\text{}}\end{array}$

3. TomLikesPhysics Group Title

$V=\int\limits_{?}^{?}f(x,y,z)dV$ $f(x,y,z)=xy , x,y \ge , x^2+y^2+z^2\le1$

4. TomLikesPhysics Group Title

Somehow I felt sphere and used the jacobi-determinant to rewrite the integral and to get V=4/3 * pi but to rewrite it I used a z component which f does not have so I am kind of clueless.

5. TomLikesPhysics Group Title

$f(x,y,z)=xy$ $x,y \ge0$ $x^2+y^2+z^2$

6. TomLikesPhysics Group Title

$x^2+y^2+z^2 \le 1$

7. TomLikesPhysics Group Title

Those are the conditions I have.

8. amistre64 Group Title

well, it does appear to be an 4th a sphere then

9. amistre64 Group Title

|dw:1353510305015:dw|

10. TomLikesPhysics Group Title

Here is my calculation.

11. TomLikesPhysics Group Title

f(x,y,z)=xy so it is a surface I thought and if I integrate that I get a 4-dim sphere?

12. amistre64 Group Title

yeah, ill have to review this later, work is calling nice work on that jacobian tho

13. TomLikesPhysics Group Title

yeah but I think nobody ask for it.^^

14. TomLikesPhysics Group Title

I mean I guess I don´t need that jacobi-det. here, right?

15. TuringTest Group Title

I'm confused as to what your final integral is....

16. TomLikesPhysics Group Title

4/3*pi

17. TomLikesPhysics Group Title

After I got that I checked the Jacobi-Det. to make sure I did not misread something on Wikipedia.

18. TuringTest Group Title

well, let me try it, but I just don't quite see the set up

19. TomLikesPhysics Group Title

The set up is everything on the left side of the picture above the line with the 1 in a circle underneath it.

20. TuringTest Group Title

I seem to get zero

21. TuringTest Group Title

$x=\rho\sin\phi\cos\theta$$y=\rho\sin\phi\sin\theta$$z=\cos\phi$$dV=\rho^2\sin\phi d\theta d\rho d\phi$$f(x,y,z)=xy=\rho^2\sin^2\phi\sin\theta\cos\theta$$\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{2\pi}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi$$=\int_0^\pi\int_0^1\rho^4\sin^3\phi\left[\left.\frac{\sin^2\theta}2\right|_0^{2\pi}\right] d\rho d\phi=0$ did I make a mistake? I hate it when I get zero...

22. TuringTest Group Title

that should be $$z=\rho\cos\theta$$ of course

23. TomLikesPhysics Group Title

I think the theta integral goes from 0 to pi. The phi integral goes from 0 to 2pi.

24. TomLikesPhysics Group Title

But that does not change a thing I think...

25. TomLikesPhysics Group Title

Ups, my mistake, you change the angles.

26. TuringTest Group Title

yeah I switched your theta for my hi just cuz that's how I usually do it

27. TuringTest Group Title

my phi*

28. TomLikesPhysics Group Title

So perhaps sphericalcoordinates do not help with this problem?

29. TomLikesPhysics Group Title

but I ended up with 4/3*pi when I used the Jacobi-Det.

30. TomLikesPhysics Group Title

but I also threw the function f=xy right out of the window and did not include it in the calculation...

31. TomLikesPhysics Group Title

Perhaps my calculation on the picture which comes right after the set up is right? The block with the 1, I mean.

32. TuringTest Group Title

you can't just throw out the thing you are integrating, and by the 'jacobi-det' I can only presume you mean the jacobian dV in spherical coordinates $$\rho^2\sin\phi$$ which I used as well the change in coordinates should not affect the answer if we do it right, which I think we did, so I think the answer is zero 4/3pi is just the volume of the shpere over which we are integrating, but the integral itself is zero

33. TomLikesPhysics Group Title

We are neglecting the part where it is mandatory that x,y are greater or equal to zero. The x in terms of sphericalcoordinates can be negative but it shouldn´t be. I think we must somehow work that and the condition that the sum of the squares is 1 into the integral.

34. TuringTest Group Title

oh, I missed that :P if x and y are greater than zero, we are in the first quadrant of the xy-plane, so theta varies from 0 to pi/2$\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{\frac\pi2}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi$

35. TomLikesPhysics Group Title

So the volume or whatever we just calculated is just 4/30? Did I do that right? Thx a lot so far Turing Test.

36. TuringTest Group Title

$\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{\frac\pi2}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi$$=\frac12\int_0^\pi\int_0^1\rho^4\sin^3\phi d\rho d\phi=\frac1{10}\int_0^\pi \sin^3\phi d\phi=$$\frac1{10}\int_0^\pi\sin\phi-\cos^2\phi\sin\phi d\phi=\frac1{10}\left(\left.-\cos\phi+\frac{\cos^3\phi}3\right|_0^\pi\right)$$=\frac1{10}(\frac23-(-\frac23))=\frac2{15}$so yes, I guess that's right. btw the thing that we found is the integral of the function f(x,y,z)=xy over the given volume of a quarter-sphere, E.

37. TomLikesPhysics Group Title

Since is has something to do with a sphere isn´t it weird that we got 2/15 and not something with pi?

38. TuringTest Group Title

No, not really. Just because the region you are integrating in is a sphere doesn't meant you should expect pi to show up. If you think about any integral in either polar, cylindrical, or spherical coordinates, if theta shows up in the integrand as the argument of a trig function, then evaluation ion the d(theta) integration will cause he angle (which is where the pi would come from, of course) converted into whatever the resulting trig function results in. example in polar coordinates for simplicity:$\int_0^1\int_0^{\pi}rdrd\theta=\frac\pi2$(with no integrand, this integral just represents the area of the region of integration; a circle with radius 1) However if we wanted to integrate $$y=r\sin\theta$$ *over* that region of a half-circle we get$\int_0^1\int_0^\pi r^2\sin\theta drd\theta=\frac23$Notice the $$\pi$$ got swallowed up by the fact that it was evaluated as the argument of a trig function, which is very common an in no way unusual in these situations.

39. TomLikesPhysics Group Title

Ok. Thank you Turing Test for your durability and stamina. You made the whole thing a lot clearer.

40. TuringTest Group Title

Happy I could help :)