Looking for advice/direction on the following query please.
Ok, here is the problem:
If we have
$$ \vec{T} = \frac{\mu IbN}{4\pi L}\int_{-L/2}^{L/2}C\int_{0}^{2\pi}\frac{\cos(\theta)}{p^3}d\theta dz $$
Where
$$ C = Z-z \\ p^{-3} = (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))(X^2+C^2)^{-3/2} $$
Now, after integration the result I should have is
$$ T = \frac{\mu}{4\pi}\frac{M}{L}X\left[(X^2 + C^2)^{-3/2} \right]_{-L/2-Z}^{L/2-Z} $$
Where
$$ M = Ib^2\pi N $$
I would appreciate if someone could verify that I'm performing the Math correctly and if not, where am I going wrong...
Ok, now here is my attempt to obtain the solution:
Step 1 is to substitute in for $ p^3 $:
$$\begin{align} \int_{0}^{2\pi}\frac{\cos(\theta)}{p^3}d\theta & = \int_{0}^{2\pi}\frac{\cos(\theta)}{(1 + \frac{3bX}{X^2 + C^2} \cos(\theta))(X^2+C^2)^{-3/2}}d\theta \\ & = \frac{1}{(X^2+C^2)^{-3/2}}\int_{0}^{2\pi}\frac{\cos(\theta)}{(1 + \frac{3bX}{X^2 + C^2} \cos(\theta))}d\theta \\ & = \frac{1}{(X^2+C^2)^{-3/2}}\frac{2\pi (C^2 +X^2)}{3Xb} \\ & = \frac{2\pi (C^2 + X^2)^{5/2}}{3Xb} \end{align}$$
Step 2 - Substitute into the double integral above:
$$ \int_{-L/2}^{L/2}C\frac{2\pi (C^2 + X^2)^{5/2}}{3Xb}dz $$
Now, as previously stated $ C = Z - z $, therefore:
$$\begin{align} \int_{-L/2}^{L/2}C\frac{2\pi (C^2 + X^2)^{5/2}}{3Xb}dz & = \int_{-L/2}^{L/2}(Z-z)\frac{2\pi ((Z-z)^2 + X^2)^{5/2}}{3Xb}dz \\ & = \frac{2\pi}{3Xb}\int_{-L/2}^{L/2}(Z-z)((Z-z)^2 + X^2)^{5/2}dz \end{align}$$
Ok, now this is where I could do with a sanity check...
1) Am I correct with my Math so far?
2) To perform the integration above, what is the best method to use?
3) Is there a way to simplify the integral at all?
I cannot see how to get from here:
$$ \frac{\mu IbN}{4\pi L}\frac{2\pi}{3Xb}\int_{-L/2}^{L/2}(Z-z)((Z-z)^2 + X^2)^{5/2}dz $$
to here:
$$ T = \frac{\mu}{4\pi}\frac{M}{L}X\left[(X^2 + C^2)^{-3/2} \right]_{-L/2-Z}^{L/2-Z} $$
From the given solution we have:
$$\begin{align} \frac{\mu}{4\pi}\frac{M}{L} & = \frac{\mu Ib^2\pi N}{4\pi L} \end{align}$$
From my Math so far we have:
$$\begin{align} \frac{\mu IbN}{4\pi L}\frac{2\pi}{3Xb} & = \frac{2\mu Ib\pi N}{12\pi LXb} \\ & = \frac{\mu I\pi N}{6\pi LX} \end{align}$$
Something seems sadly amiss - any advice?
For the Integral, I cannot see how to get from:
$$ \int_{-L/2}^{L/2}(Z-z)((Z-z)^2 + X^2)^{5/2}dz $$
to:
$$ \left[(X^2 + C^2)^{-3/2} \right]_{-L/2-Z}^{L/2-Z} $$
Not really looking for the answers, just some pointers as to where I maybe going wrong. In particular the correct method of solution for the Integral above.
Thankyou.
Ok, continuing on from the comments recieved...
As noted:
$$\begin{align} \frac{\cos(\theta)}{p^3} & = \cos(\theta)*p^{-3} \\ & = \cos(\theta) * (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))(X^2+C^2)^{-3/2} \end{align} $$
So if I revisit the first integral we have:
Step 1:
$$\begin{align} \int_{0}^{2\pi}\frac{\cos(\theta)}{p^3}d\theta & = \int_{0}^{2\pi}\cos(\theta) * (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))(X^2+C^2)^{-3/2}d\theta \\ & = (X^2+C^2)^{-3/2}\int_{0}^{2\pi}\cos(\theta) * (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))d\theta \\ & = (X^2+C^2)^{-3/2} \frac{3\pi Xb}{C^2+X^2} \\ & = (3\pi Xb)(C^2+X^2)^{-5/2} \end{align}$$
Step 2 is to apply this within the second integral, so we have:
$$\begin{align} \int_{-L/2}^{L/2}3\pi XbC(C^2+X^2)^{-5/2}dz & = 3\pi Xb\int_{-L/2}^{L/2}C(C^2+X^2)^{-5/2}dz \\ & = 3\pi Xb\int_{-L/2}^{L/2}(Z-z)((Z-z)^2+X^2)^{-5/2}dz \end{align}$$
Now, is there something 'special' about the integral above? Mathcad will not solve this - is there some elliptical integral involved in this?
Any advice/pointers appreciated.
Ok, final update and hopefully I will show how to get the result I was looking for...
Step 1:
$$\begin{align} \int_{0}^{2\pi}\frac{\cos(\theta)}{p^3}d\theta & = \int_{0}^{2\pi}\cos(\theta) * (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))(X^2+C^2)^{-3/2}d\theta \\ & = (X^2+C^2)^{-3/2}\int_{0}^{2\pi}\cos(\theta) * (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))d\theta \\ & = (X^2+C^2)^{-3/2} \frac{3\pi Xb}{C^2+X^2} \\ & = (3\pi Xb)(C^2+X^2)^{-5/2} \end{align}$$
Step 2 is to apply this within the second integral, so we have:
$$\begin{align} \int_{-L/2}^{L/2}3\pi XbC(C^2+X^2)^{-5/2}dz & = 3\pi Xb\int_{-L/2}^{L/2}C(C^2+X^2)^{-5/2}dz \\ & = 3\pi Xb\left[\frac{8}{3*(L^2+4*X^2-4*ZL+4*Z^2}^{-3/2}-\frac{8}{3*(L^2+4*X^2+4*ZL+4*Z^2}^{-3/2}\right] \\ & = \frac{3\pi Xb}{3}\left[\left[(Z-\frac{L}{2})^2+X^2\right]^{-3/2} - \left[(Z+\frac{L}{2})^2+X^2\right]^{-3/2}\right] \\ & = \pi Xb\left[\left[(Z-z)^2+X^2)\right]^{-3/2}\right]_{-L/2-Z}^{L/2-Z} \end{align}$$
The original result required was:
$$\begin{align} T & = \frac{\mu}{4\pi}\frac{M}{L}X\left[(X^2 + C^2)^{-3/2} \right]_{-L/2-Z}^{L/2-Z} \\ & = \frac{\mu Ib^2\pi N}{4\pi L}X\left[(X^2 + C^2)^{-3/2} \right]_{-L/2-Z}^{L/2-Z} \end{align}$$
And we now have:
$$\begin{align} T & = \frac{\mu IbN}{4\pi L}\int_{-L/2}^{L/2}C\int_{0}^{2\pi}\frac{\cos(\theta)}{p^3}d\theta dz \\ & = \frac{\mu Ib^2\pi NX}{4\pi L}\left[\left[(Z-z)^2+X^2)\right]^{-3/2}\right]_{-L/2-Z}^{L/2-Z} \\ & = \frac{\mu Ib^2\pi N}{4\pi L}X\left[(X^2 + C^2)^{-3/2} \right]_{-L/2-Z}^{L/2-Z} \end{align}$$
Which I think is correct?
The main error in your calculation I see, is that of your definition and implementation of $p$. You define that $$p^{-3} = (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))(X^2+C^2)^{-3/2}$$ and you have $\frac{\cos \theta}{p^3}$ which is equivalent to saying $\cos\theta \cdot p^{-3}$. When you go to substitute in $p$, you still divide $\cos\theta$ by $p^{-3}$ rather than multiplying it. As such, you will result in this:
$$ \frac{\cos\theta}{p^3} = \cos\theta\cdot p^{-3}\\ p^{-3} = (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))(X^2+C^2)^{-3/2} \\ \text{Therefore, when you substitute in }p \text{ you get: }\\ \cos\theta\cdot (1 + \frac{3bX}{X^2 + C^2} \cos(\theta))(X^2+C^2)^{-3/2} $$