Volume of Revolution: Find Vol. Bounded by y=6-2x-x2, y=x+6 About x=1

[Aurin]

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.

 

[Mark44]

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

The radius, which you are calling p(x), is 1 - x, not x + 1. For example, at the point (-3, 3), the distance to the line x = 1 is 1 - (-3) = 4. Similarly, at the point (0, 6), the distance to the line x = 1 is 1 - 0 = 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.

 

[Aurin]

Oh man.. haha, what a ridiculous mistake.

I had looked at many examples to see if my p(x) followed the rules, and one of them did indeed add 1 to x, so it made so much sense to try it here, but it's not the same case. Negatives are my weakness. =(

Thank you very much!