Base bounded by y = 1 - x^2 and y = 0. Cross-sections perpendicular to the x-axis are squares with side equal to the height of the base (1 - x^2). Volume:

\tfrac{16}{15} Side = 1 - x^2. Area = (1 - x^2)^2. V = \int{-1}^{1}(1 - x^2)^2\,dx = 2\int0^1(1 - 2x^2 + x^4)\,dx = 2(1 - 2/3 + 1/5) = 2(8/15) = 16/15.

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Volumes with Cross Sections

AP Calculus AB· difficulty 3/5

y = 1 - x^2

A
$\tfrac{16}{15}$
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B
$\tfrac{4}{3}$
C
$1$
D
$\tfrac{2}{3}$

Explanation

= 1 - x^2

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