- #1

calchelp

- 11

- 0

the solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. the cross-sections perpendicular to the axis on the interval 0 ≤ x ≤ 4 are squares whose diagonals run from the parabola y = - sqrt x to the parabola y = sqrt x.

ok so...

the base of the square is a.

the diagonal is 2a

^{2}.

the diagonal is 2 sqrt x.

so a

^{2}, the area of the square, is sqrt x.

therefore,

{integral} from 0 to 4 of sqrt x

antidifferentiate and you get 2/3 x

^{3/2}

plug in 4 and 0

final answer: 16/3

book says: 16

where did i mess up?