Find the volume of revolution of the region generated,
bounded by graph $\displaystyle y = 6 - 2x -x^2$
and the indicated line $\displaystyle y = x + 6.$
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)
$\displaystyle V = \pi \int\limits_{ - 3}^0 {\left[ {\left( {6 - 2x - x^2 } \right)^2 - \left( {x + 6} \right)^2 } \right]} .dx \hfill \\$
$\displaystyle = \pi \int\limits_{ - 3}^0 {\left[ {\left( {x^4 + 4x^3 - 8x^2 - 24x + 36} \right) - \left( {x^2 + 12x + 36} \right)} \right]} .dx \hfill \\$
$\displaystyle = \pi \int\limits_{ - 3}^0 {\left[ {x^4 + 4x^3 - 9x^2 - 36x} \right]} .dx \hfill \\$
$\displaystyle = \pi \left[ {\frac{{x^5 }}
{5} + x^4 - 3x^3 - 18x^2 } \right]_{ - 3}^0 \hfill \\$
$\displaystyle = \pi \left[ {\left( {\frac{{\left( { - 3} \right)^5 }}
{5} + \left( { - 3} \right)^4 - 3\left( { - 3} \right)^3 - 18\left( { - 3} \right)^2 } \right) - \left( {0 + 0 - 0 - 0} \right)} \right] \hfill \\$
$\displaystyle = \pi \left[ {\frac{{ - 243}}
{5} + 81 + 81 + 162} \right] \hfill \\$
$\displaystyle = \pi \left[ {\frac{{ - 243}}
{5} + 324} \right] \hfill \\$
$\displaystyle = \frac{{1377\pi }}
{5} \hfill \\ $
This answer seems too big to me. Is it correct?
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 ...
one error in the integral evaluation below ... you did the FTC evaluation backwards, should be F(0) - F(-3)$\displaystyle = \pi \left[ {\frac{{x^5 }}
{5} + x^4 - 3x^3 - 18x^2 } \right]_{ - 3}^0 \hfill \\$
sign error in this next step ... +162 ?$\displaystyle = \pi \left[ {\left( {\frac{{\left( { - 3} \right)^5 }}
{5} + \left( { - 3} \right)^4 - 3\left( { - 3} \right)^3 - 18\left( { - 3} \right)^2 } \right) - \left( {0 + 0 - 0 - 0} \right)} \right] \hfill \\$
$\displaystyle = \pi \left[ {\frac{{ - 243}}
{5} + 81 + 81 + 162} \right] \hfill \\$