Hepl...integration probelm ....math...please ....

[cyw1984]

a, b,h is constant
how to solve it?
thx??

 

[cedance]

I dont see any difficulty here. or is there?

first integrate for the dr integral, where z is constant. add and subtract z² and split the integral and u can solve for the dr. then r is constant and solve for z. did u try it this way?

cedance.

 

[v_c]

The inner integral evaluates to

\[\int \frac{r^2}{\sqrt[3]{r^2+z^2}} \, dr \,\,\,= \,\, \rm{arcsinh}(\frac{r}{z}) \,\, - \,\, \frac{r}{\sqrt{r^2+z^2}}\]

I am running into trouble with the outer integral once I apply the upper and
lower limits to the result above.

Best regards,
v_c

 

[cedance]

Hi,

Ok, finally, I solved the problem and posting the solution here. It took me 5 minutes to complete it and 1 hour to type it in LATEX! you owe me one!

i have checked the solution with a couple of sample inputs, and turned out to be right. however, in case you find it wrong, i advise you to go thro the solution and check for the mistakes for i hope the method to be right, only some calculation errors shud be there, IF any.

cedance.

PS: v_c, the solution to your integral is right, but you have typed the LHS wrong i guess, its to the power 3/2 not 1/3. u have written the cube root expression??

 

[v_c]

cedance,

that's a good catch! Yes it was supposed to be to the power of 3/2.
I was using the TeX equations and got a little careless.

Best regards,
v_c