Question
can anyone help me with the last problem # 14
Find the volume of the given solid generated by revolving the regions bounded by the graphs of the equations about the given lines.
14. y=6-2x-x^2, y=x+6
a. the x-axis
b. the line y=3
Find the volume of the given solid generated by revolving the regions bounded by the graphs of the equations about the given lines.
14. y=6-2x-x^2, y=x+6
a. the x-axis
b. the line y=3
Answers
The graphs intersect at (-3,3) and (0,6)
Using vertical strips is easiest, since the upper and lower boundaries do not change. That means we integrate on x.
For revolving around the x-axis, that means using discs (washers) of thickness dx, so
v = ∫[-3,0] π(R^2-r^2) dx
where R=6-2x-x^2 and r=x+6
v = ∫[-3,0] π((6-2x-x^2)^2-(x+6)^2) dx = 243π/5
To revolve around y=3, the same idea is used, but each radius is now y-3 instead of y.
Using vertical strips is easiest, since the upper and lower boundaries do not change. That means we integrate on x.
For revolving around the x-axis, that means using discs (washers) of thickness dx, so
v = ∫[-3,0] π(R^2-r^2) dx
where R=6-2x-x^2 and r=x+6
v = ∫[-3,0] π((6-2x-x^2)^2-(x+6)^2) dx = 243π/5
To revolve around y=3, the same idea is used, but each radius is now y-3 instead of y.
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