Integrate y = 2pi Int dz int(1-r^2)*r dr: Solution

[maluta]

y = 2pi Int dz int(1-r^2)*r dr and the limits are (0,1) (note int = Integral)

I was trying to integrate it by parts for the second part and I think I am not doing it correctly. Which way can I apply for it.

 

[quasar987]

Try a change of variable.

 

[HallsofIvy]

y = 2pi Int dz int(1-r^2)*r dr and the limits are (0,1) (note int = Integral)

I was trying to integrate it by parts for the second part and I think I am not doing it correctly. Which way can I apply for it.

Which limits are 0, 1? Both? If this is
2\pi \int_0^1 dz \int_0^1 (1- r^2)r dr
I see no reason to integrate by parts. The first integral is
\int_0^1 dz= \left[ z\right]_0^1= 1
and the second is
\int_0^1 (1-r^2)r dr
Let u= 1-r^2. Then du= -2r dr. When r= 0, u= 1 and when r= 1, u= 0.
\int_1^0 u(-(1/2)du= \left{(-1/2)(u^2/2)\right]_1^0= (-1/2)(0- 1)= 1/2[/itex]<br /> 2\pi\int_0^1 dz \int_0^1 (1-r^2)r dr= 2\pi (1)(1/2)= \pi

 

[quasar987]

(I assumed the OP meant r/(1-r^2) for the integrand. Otherwise, there is no need to use any "integration technique" for this integral.)