- #1
Aurin
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Homework Statement
Find the volume of the solid generated by revolving the region bounded by the given graph about the indicated line using any method.
y=6-2x-x2
y=x+6
About the axis x=1
Homework Equations
Shell method:
2*pi * INT[ p(x) * h(x) ] dx
The Attempt at a Solution
First, I attempted latex, even using your beginner's guide, and it showed equations that didn't even look partially like what I input, so forgive me. =(
Second, I attempted the shell method because the disk method would require changing the equations into terms of y, and this seemed simple enough.
I've graphed the equations, and determined the following:
The two lines intersect at (-3,3) and (0,6) so the boundaries are a=-3 and b=0. The region is bound in the second quadrant, thus p(x) is the distance from the y-axis plus the distance of the axis of rotation from the y-axis. h(x) is the smaller equation subtracted from the larger equation.
p(x) = x+1
h(x) = 6-2x-x2 - (x+6) = -3x-x2
2*pi * INT[(x+1)(-3x-x2)]dx Plugged in p(x) and h(x)
2*pi * INT(-x3-4x2-3x)dx FOIL
2*pi * (-x4/4 - 4x3/3 - 3x2/2) from -3 to 0. Integrated
If you plug in 0 you receive 0, thus
2*pi * [-(-81/4 + 108/3 - 27/2)] Carefully substituting -3 into the equation.
2*pi * (81/4 - 108/3 + 27/2) Multiplied the -1
2*pi * (243/12 - 432/12 + 162/12) Found a common denominator
2*pi * (-27/12) Simplified
-9pi/2 Reduced and Simplified more
So my final answer is a negative volume. I've looked at this carefully time and time again, and the result is always negative. I realize this is probably difficult to work with, considering the lack of latex and no visual, but I appreciate any input.