How Do I Correctly Calculate Volume for a UVZ Coordinate System?

sedaw
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attached herewith : the problem with my attempt to solve it .
 

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1. What is "an" in "cot ant"?

2. How did you get the limits of integration? Specifically, what are the limits of integration on u and v?
 
HallsofIvy said:
1. What is "an" in "cot ant"?

2. How did you get the limits of integration? Specifically, what are the limits of integration on u and v?

cotan(t) = cos(t)/sit(t)

the limits of integration on u & v simply received from the projection of the region bounded by V on the UV plane -> you can see it circle with radius R=1 .

the limits of integration on z:

you can see in sketch : 0<=z<=U+V (of course just for u & v that in region of the projection )
TNX ...
 
Yeah, the ant thing . . . not a good way to start sedaw. Did you just enter "cotant" in latex and it partitioned it that way? Just use \cot. But even worst, that picture looks nothing like what I think your V is. Guess the J is Jacobian. But I wouldn't try to change variables until I first tried to solve it the old-fashioned way: just use x, y, and z. Oh, but I would also try to plot a real-looking picture first. Is the volume that part under the paraboloid z=x^2+y^2[/tex] contained by the transparent &quot;square-tube&quot; of x^4+y^4=1 under the paraboloid in the plot below?<br /> <br /> If so, can you see how to arrive at:<br /> <br /> V=8\int_0^{1/2^{1/4}}\int_x^{\sqrt[4]{1-x^4}}\left(x^2+y^2\right) dydx<br /> <br /> That comes out to about 2.22. Would be nice if you already know the answer to see at least if this is on the right track and I realize that&#039;s a tough integral to solve symbolically, but it get&#039;s you grounded at least with something concrete to work with and then perhaps you can use that integral to then change variables. Maybe not though. It&#039;s just a suggestion.
 

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ok.. thank you very much i am appreciate your help.
by the way my picture is V for the uvz axis and not xyz .
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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