Triple integration continued
Gentlemen (PH and galactus)
Thank you very much.
Perfect Hacker, in reply 3, you mean let u = sinx and not u = cosx. If we let u =sinx, then du = cosxdx, that way we can integrate (1 - u^2)^2du.
To both of you,
I also did the question in rectangular as in reply 4. This makes sense as I got the same limits of integration. I get a final answer of (3pi -8)/(12(2)^1/2) = 0.0840 approx.
I don't understand changing the variables and applying Jacobian.
I have only done double integration for general regions as type I or II, polar coordinates, triple integration in general regions type I or II, circular coordinates. I have not got to spherical coordinates yet so anything past that point is above my head. However, you have made the Jacobian clear in reply 5 and I am able to understand it and see that it is a much easier integral to work with. I pressume that the rectangular and Jacobian methods should give the same volume right?!
Galactus in reply 7 you agree with PH that the best method is rectangular. I also agree with your cylindrical as that is what I have. You seem to integrate with respect to theta from 0 to 2pi instead of 0 to pi/2, is this a typo?
I have used quickmath.com to evaluate the iterated integral using cylindrical coordinates, I get to the last integral with respect to theta which I will name x for convenience right now and I get the integral expression of:
1/6[(3/cos^2(x) + 1) - sin(x)/(cos^2(x) + 1)^2 from x = 0 to x = pi/2. Remember I am using x in place for theta for convenience!
The computer program gives me a final answer in terms of many arctan functions which is simplified to an answer of 2.409, much higher than the answer I obtained from rectangular coordinates of 0.0840!
I will calculate the Jacobian method and get an answer but I need two clarifications first: Since it is the first octant we are integrating from 0 to pi/2 and not from 0 to 2pi, correct?! Also when integrating r - r^2sin(theta) with respect to r, sin(theta) is treated as a constant and thus we end up with r^2 - sin(theta)(r^3)/3 from r = 0 to r = 1.
I cannot help to notice that the Jacobian method simplifies the cylindrical method by doing a determinant and simplifying the expression for the upper limit of r to just r =1, right?
I really appreciate your help gentlemen. As you can see I am not slacking off but trying very hard. I am happy that your methods agree with what I am able to get.