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TomLikesPhysics
 3 years ago
I need help with an integral.
TomLikesPhysics
 3 years ago
I need help with an integral.

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \begin{array}l\color{red}{\text{I}}\color{orange}{\text{n}}\color{#e6e600}{\text{t}}\color{green}{\text{e}}\color{blue}{\text{g}}\color{purple}{\text{r}}\color{purple}{\text{a}}\color{red}{\text{t}}\color{orange}{\text{i}}\color{#e6e600}{\text{o}}\color{green}{\text{n}}\color{blue}{\text{ }}\color{purple}{\text{n}}\color{purple}{\text{e}}\color{red}{\text{e}}\color{orange}{\text{d}}\color{#e6e600}{\text{s}}\color{green}{\text{ }}\color{blue}{\text{n}}\color{purple}{\text{o}}\color{purple}{\text{ }}\color{red}{\text{h}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{p}}\color{blue}{\text{ }}\color{purple}{\text{.}}\color{purple}{\text{.}}\color{red}{\text{.}}\color{orange}{\text{.}}\color{#e6e600}{\text{ }}\color{green}{\text{✌}}\color{blue}{\text{}}\end{array} \]

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1\[V=\int\limits_{?}^{?}f(x,y,z)dV\] \[f(x,y,z)=xy , x,y \ge , x^2+y^2+z^2\le1\]

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1Somehow I felt sphere and used the jacobideterminant 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.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1\[f(x,y,z)=xy\] \[x,y \ge0\] \[x^2+y^2+z^2\]

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1\[x^2+y^2+z^2 \le 1\]

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1Those are the conditions I have.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0well, it does appear to be an 4th a sphere then

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353510305015:dw

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1Here is my calculation.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1f(x,y,z)=xy so it is a surface I thought and if I integrate that I get a 4dim sphere?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, ill have to review this later, work is calling nice work on that jacobian tho

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1yeah but I think nobody ask for it.^^

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1I mean I guess I don´t need that jacobidet. here, right?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2I'm confused as to what your final integral is....

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1After I got that I checked the JacobiDet. to make sure I did not misread something on Wikipedia.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2well, let me try it, but I just don't quite see the set up

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1The set up is everything on the left side of the picture above the line with the 1 in a circle underneath it.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2\[x=\rho\sin\phi\cos\theta\]\[y=\rho\sin\phi\sin\theta\]\[z=\cos\phi\]\[dV=\rho^2\sin\phi d\theta d\rho d\phi\]\[f(x,y,z)=xy=\rho^2\sin^2\phi\sin\theta\cos\theta\]\[\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{2\pi}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi\]\[=\int_0^\pi\int_0^1\rho^4\sin^3\phi\left[\left.\frac{\sin^2\theta}2\right_0^{2\pi}\right] d\rho d\phi=0\] did I make a mistake? I hate it when I get zero...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2that should be \(z=\rho\cos\theta\) of course

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1I think the theta integral goes from 0 to pi. The phi integral goes from 0 to 2pi.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1But that does not change a thing I think...

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1Ups, my mistake, you change the angles.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2yeah I switched your theta for my hi just cuz that's how I usually do it

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1So perhaps sphericalcoordinates do not help with this problem?

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1but I ended up with 4/3*pi when I used the JacobiDet.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1but I also threw the function f=xy right out of the window and did not include it in the calculation...

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1Perhaps my calculation on the picture which comes right after the set up is right? The block with the 1, I mean.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2you can't just throw out the thing you are integrating, and by the 'jacobidet' 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

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1We 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.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2oh, I missed that :P if x and y are greater than zero, we are in the first quadrant of the xyplane, so theta varies from 0 to pi/2\[\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{\frac\pi2}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi\]

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1So the volume or whatever we just calculated is just 4/30? Did I do that right? Thx a lot so far Turing Test.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2\[\iiint\limits_Ef(x,y,z)dV=\int_0^\pi\int_0^1\int_0^{\frac\pi2}\rho^4\sin^3\phi\sin\theta\cos\theta d\theta d\rho d\phi\]\[=\frac12\int_0^\pi\int_0^1\rho^4\sin^3\phi d\rho d\phi=\frac1{10}\int_0^\pi \sin^3\phi d\phi=\]\[\frac1{10}\int_0^\pi\sin\phi\cos^2\phi\sin\phi d\phi=\frac1{10}\left(\left.\cos\phi+\frac{\cos^3\phi}3\right_0^\pi\right)\]\[=\frac1{10}(\frac23(\frac23))=\frac2{15}\]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 quartersphere, E.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1Since is has something to do with a sphere isn´t it weird that we got 2/15 and not something with pi?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2No, 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 halfcircle 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.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.1Ok. Thank you Turing Test for your durability and stamina. You made the whole thing a lot clearer.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2Happy I could help :)
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