Explanation of the solution to definite integral

[jn1998]

Could someone please explain, in a series of steps if possible, how to get to the solution of this definite integral please? I am mainly confused by the stuff in the square brackets. Any help would be much appreciated, thanks!

 

[skeeter]

I don’t agree with the given evaluation of the definite integral ...

[MATH]\pi re^2 \int_{-rp}^{rp} 1 - \dfrac{z^2}{rp^2} \, dz[/MATH]
[MATH]\pi re^2 \bigg[z - \dfrac{z^3}{3rp^2} \bigg]_{-rp}^{rp}[/MATH]
[MATH]\pi re^2 \bigg[\left(rp - \dfrac{r^3p^3}{3rp^2} \right) - \left(-rp + \dfrac{r^3p^3}{3rp^2} \right) \bigg] [/MATH]
[MATH]\pi re^2 \left(2rp - \dfrac{2r^2p}{3}\right)[/MATH]
[MATH]2\pi r^2e^2 \left(p - \dfrac{rp}{3} \right) [/MATH]

 

[jn1998]

Thanks so much for the reply. However, the answer is definitely correct. Apologies, I wrote out the integral and answer incorrectly. This is the modified integral and answer:

 

[Deleted member 4993]

Thanks so much for the reply. However, the answer is definitely correct. Apologies, I wrote out the integral and answer incorrectly. This is the modified integral and answer:

View attachment 21621

Response #2 above is correct when you replace "rp" by "r_p" (according to your revised problem statement.

 

[skeeter]

Thanks so much for the reply. However, the answer is definitely correct. Apologies, I wrote out the integral and answer incorrectly. This is the modified integral and answer:

View attachment 21621

So, both p and e are subscripts ...

[MATH]\pi r_e^2 \int_{-r_p}^{r_p} 1 - \dfrac{z^2}{r_p^2} \, dz[/MATH]
[MATH]\pi r_e^2 \bigg[z - \dfrac{z^3}{3r_p^2}\bigg]_{-r_p}^{r_p} [/MATH]
[MATH]\pi r_e^2 \bigg[\left(r_p - \dfrac{r_p^3}{3r_p^2}\right) - \left(-r_p + \dfrac{r_p^3}{3r_p^2} \right) \bigg] = \pi r_e^2 \cdot \dfrac{4 r_p}{3}[/MATH]