Question

can someone help me in calculus 2?

Answer 1

oops # 13

Answer 2

@nincompoop

Answer 3

you forget about pi as the factor in the last few lines

Answer 4

ahhh leh me see where

Answer 5

oh okay so then

Answer 6

yes, you're ok, if you put in the pi
\[ 2\pi \left(\frac{4\pi}{3} - \sqrt{3} \right) \\
\frac{8\pi^2}{3} - 2\pi \sqrt{3}\]