volumes by cross - section lecture participation: problem 2\n(2 points)\nfind the volume of the solid s…

volumes by cross - section lecture participation: problem 2\n(2 points)\nfind the volume of the solid s described below.\nthe base of s is the region enclosed by the parabola ( y = 4 - x^{2} ) and the x - axis. cross - sections perpendicular to the y - axis are squares.\nvolume = \npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.

volumes by cross - section lecture participation: problem 2\n(2 points)\nfind the volume of the solid s described below.\nthe base of s is the region enclosed by the parabola ( y = 4 - x^{2} ) and the x - axis. cross - sections perpendicular to the y - axis are squares.\nvolume = \npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.

Answer

Answer:

$32$

Explanation:

Step1: Express (x) in terms of (y)

Given (y = 4 - x^{2}), we can rewrite it as (x=\pm\sqrt{4 - y}). The length of the side of the square cross - section perpendicular to the (y) - axis is (s = 2\sqrt{4 - y}) (since the distance between (x =-\sqrt{4 - y}) and (x=\sqrt{4 - y}) is (2\sqrt{4 - y})).

Step2: Find the area of the cross - section

The area of a square (A(y)=s^{2}). Substituting (s = 2\sqrt{4 - y}) into the formula, we get (A(y)=(2\sqrt{4 - y})^{2}=4(4 - y)=16-4y).

Step3: Determine the limits of integration

When (x = 0), (y = 4) (from the equation (y = 4 - x^{2})). When the solid is on the (x) - axis ((y = 0)). So the limits of integration for (y) are from (y = 0) to (y = 4).

Step4: Calculate the volume using the formula (V=\int_{a}^{b}A(y)dy)

We use the formula (V=\int_{0}^{4}(16 - 4y)dy). Integrating term - by - term: (\int(16-4y)dy=16y-2y^{2}+C). Evaluating the definite integral (\left[16y-2y^{2}\right]_{0}^{4}=(16\times4-2\times4^{2})-(0)). [ \begin{align*} &16\times4-2\times16\ =&64 - 32\ =&32 \end{align*} ]