# Volume of Revolution

• Apr 3rd 2009, 05:27 PM
Shyam
Volume of Revolution
Find the volume of revolution of the region generated,

bounded by graph $y = 6 - 2x -x^2$

and the indicated line $y = x + 6.$
• Apr 3rd 2009, 05:34 PM
skeeter
Quote:

Originally Posted by Shyam
Find the volume of revolution of the region generated,

bounded by graph $y = 6 - 2x -x^2$

and the indicated line $y = x + 6.$

rotated about a line? which one?
• Apr 3rd 2009, 06:42 PM
Shyam
Quote:

Originally Posted by skeeter
rotated about a line? which one?

rotated about line y = x+6
• Apr 3rd 2009, 07:34 PM
Shyam
Quote:

Originally Posted by Shyam
rotated about line y = x+6

here's the work I did:
After solving the equations of parabola and line, I found their point of intersection.
The point of intersection of parabola and line are (-3, 3) and (0, 6)

$V = \pi \int\limits_{ - 3}^0 {\left[ {\left( {6 - 2x - x^2 } \right)^2 - \left( {x + 6} \right)^2 } \right]} .dx \hfill \\$

$= \pi \int\limits_{ - 3}^0 {\left[ {\left( {x^4 + 4x^3 - 8x^2 - 24x + 36} \right) - \left( {x^2 + 12x + 36} \right)} \right]} .dx \hfill \\$

$= \pi \int\limits_{ - 3}^0 {\left[ {x^4 + 4x^3 - 9x^2 - 36x} \right]} .dx \hfill \\$

$= \pi \left[ {\frac{{x^5 }}
{5} + x^4 - 3x^3 - 18x^2 } \right]_{ - 3}^0 \hfill \\$

$= \pi \left[ {\left( {\frac{{\left( { - 3} \right)^5 }}
{5} + \left( { - 3} \right)^4 - 3\left( { - 3} \right)^3 - 18\left( { - 3} \right)^2 } \right) - \left( {0 + 0 - 0 - 0} \right)} \right] \hfill \\$

$= \pi \left[ {\frac{{ - 243}}
{5} + 81 + 81 + 162} \right] \hfill \\$

$= \pi \left[ {\frac{{ - 243}}
{5} + 324} \right] \hfill \\$

$= \frac{{1377\pi }}
{5} \hfill \\$

This answer seems too big to me. Is it correct?
• Apr 4th 2009, 08:28 AM
skeeter
Quote:

Originally Posted by Shyam
rotated about line y = x+6

what you have shown is the work required for a rotation of the region about the x-axis, not about the line y = x+6.

this is correct ...

Quote:

$= \pi \left[ {\frac{{x^5 }}
{5} + x^4 - 3x^3 - 18x^2 } \right]_{ - 3}^0 \hfill \\$

one error in the integral evaluation below ... you did the FTC evaluation backwards, should be F(0) - F(-3)

Quote:

$= \pi \left[ {\left( {\frac{{\left( { - 3} \right)^5 }}
{5} + \left( { - 3} \right)^4 - 3\left( { - 3} \right)^3 - 18\left( { - 3} \right)^2 } \right) - \left( {0 + 0 - 0 - 0} \right)} \right] \hfill \\$

sign error in this next step ... +162 ?

Quote:

$= \pi \left[ {\frac{{ - 243}}
{5} + 81 + 81 + 162} \right] \hfill \\$