# Help with an integral

• Sep 2nd 2009, 01:48 AM
guest
Help with an integral
Hi.

Can anybody help me with the solution of the next integral:

$\displaystyle \int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi$

where $\displaystyle \theta=\theta(\phi)$.

The problem is that don't know the relation between $\displaystyle \theta$ and $\displaystyle \phi$.
• Sep 2nd 2009, 01:54 PM
skeeter
Quote:

Originally Posted by guest
Hi.

Can anybody help me with the solution of the next integral:

$\displaystyle \int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi$

where $\displaystyle \theta=\theta(\phi)$.

The problem is that don't know the relation between $\displaystyle \theta$ and $\displaystyle \phi$.

looks somewhat like an arc length integral for a polar function ... in what context was this integral set up?
• Sep 2nd 2009, 02:38 PM
JeffN12345
lets try this: call the value of the definate integral f, square it and evaluate it as a double integral. the square root of the double integral should be f also lets pretend that theta and phi are independant since we havent got a relationship between them

$\displaystyle f=\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi$
$\displaystyle f^2=(\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi)^2$
$\displaystyle f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta(\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2) (d \phi)^2 $$\displaystyle f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta\sin^2 \thetad\phi^2+\int_{\phi_0}^{\phi}\int_{\phi_0}^{\ phi}\cos^2\thetad\theta^2 \displaystyle f^2=\int_{\phi_0}^{\phi}(\phi-\phi_0)\cos^2\theta\sin^2\theta*d\phi+\int_{\phi_0 }^{\phi}\frac{2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0}{4}d\theta \displaystyle f^2=(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4} finally \displaystyle f= \sqrt{(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}} i cannot guarentee this is the answer youre looking for, however this is the only way i was able to integrat the function \displaystyle e^{-x^2} good luck!! • Sep 2nd 2009, 09:56 PM guest Quote: Originally Posted by skeeter looks somewhat like an arc length integral for a polar function ... in what context was this integral set up? This is a part of the formula for the speed down an unknown path on a sphere with friction. • Sep 2nd 2009, 10:17 PM guest Quote: Originally Posted by JeffN12345 \displaystyle f=\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi \displaystyle f^2=(\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi)^2 \displaystyle f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta(\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2) (d \phi)^2$$\displaystyle f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta\sin^2 \theta d\phi^2+\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi}\c os^2\theta d\theta^2$
$\displaystyle f^2=\int_{\phi_0}^{\phi}(\phi-\phi_0)\cos^2\theta\sin^2\theta*d\phi+\int_{\phi_0 }^{\phi}\frac{2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0}{4}d\theta$
$\displaystyle f^2=(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}$
finally
$\displaystyle f= \sqrt{(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}}$

Your intgrating would be okay if $\displaystyle \theta \neq \theta(\phi)$. I think that because of not knowing the relation $\displaystyle \theta = \theta(\phi)$ this integral can't be solved annalyticaly.