1)

y^2 = e^x + 1

Volume = pi* integral(0 to 1) (e^x + 1) dx

= pi* [e^x + x] (0 to 1)

= pi* (e + 1) - e^0

= pi*e units cubed

2) At y = 1 we have 1 = e^2x

ln1 = 2x

x = 0

So we integrate from 0 to 2

A = integral(0 to 2) (e^2x) dx

= [e^2x / 2] (0 to 2)

= e^4 / 2 - 1/2

= 1/2 (e^4 - 1) units squared

Edit: yeah minus the rectangle as 1729 said

Edit: I got my second answer off yahoo answers (not copy pasted but I copied their answer) rip got betrayed

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