Integrating in Spherical Coordinates

So basically what the title says;
Question 25.



I appologize for the pomegranatety picture, but it's much better than having to type all that out.


I keep getting 11pi, answer is 5pi.

 weird typed math to explain my work

int0>2pi int0>pi/3 intsecphi>2 3p^2 sinphi dp dphi dtheta
'' '' p^3 | secphi>2 dphi dtheta
'' '' (8 - sec^3phi) sinphi dphi dtheta
'' '' 8sinphi - sinsec^3phi dphi dtheta
'' '' 8sinphi - tansec^2phi dphi dtheta
u=tanphi
du=sec^2phi dphi
int u du = u^2 / 2 = tan^2phi / 2
'' -8cosphi - (tan^2 phi /2 ) |0>pi/3 dtheta
" -8cos(pi/3) - (tan(pi/3))^2 / 2 - (-8cos0 - tan^2 0 / 2) dtheta
" -4 - 3/2 - (-8 - 0) dtheta
"11/2 dtheta
11/2 "dtheta
11/2 (theta)|0>2pi
11/2(2pi)
11pi

I keep getting 11pi, answer is 5pi.

 weird typed math to explain my work

int0>2pi int0>pi/3 intsecphi>2 3p^2 sinphi dp dphi dtheta
'' '' p^3 | secphi>2 dphi dtheta
'' '' (8 - sec^3phi) sinphi dphi dtheta
'' '' 8sinphi - sinsec^3phi dphi dtheta
'' '' 8sinphi - tansec^2phi dphi dtheta
u=tanphi
du=sec^2phi dphi
int u du = u^2 / 2 = tan^2phi / 2
'' -8cosphi - (tan^2 phi /2 ) |0>pi/3 dtheta
" -8cos(pi/3) - (tan(pi/3))^2 / 2 - (-8cos0 - tan^2 0 / 2) dtheta
" -4 - 3/2 - (-8 - 0) dtheta
"11/2 dtheta
11/2 "dtheta
11/2 (theta)|0>2pi
11/2(2pi)
11pi

Double negative.

oh my god i am ridiculous