HELP!! Please, I'll give the first person to get it right a brainliest! 0.8/y =2: 3/4
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Cross sections of the volume are washers or annuli with outer radii <em>x(y)</em> + 1, where
<em>y</em> = <em>x(y) </em>² - 1 ==> <em>x(y)</em> = √(<em>y</em> + 1)
and inner radii 1. The distance between the outermost edge of each shell to the axis of revolution is then 1 + √(<em>y</em> + 1), and the distance between the innermost edge of <em>R</em> on the <em>y</em>-axis to the axis of revolution is 1.
For each value of <em>y</em> in the interval [-1, 3], the corresponding cross section has an area of
<em>π</em> (1 + √(<em>y</em> + 1))² - <em>π</em> (1)² = <em>π</em> (2√(<em>y</em> + 1) + <em>y</em> + 1)
Then the volume of the solid is the integral of this area over [-1, 3]: