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About LinkBacksFind the volume of revolution of the region generated,
bounded by graph
and the indicated line
here's the work I did:Originally Posted by Shyam
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)
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)![= \pi \left[ {\frac{{x^5 }}<br />
{5} + x^4 - 3x^3 - 18x^2 } \right]_{ - 3}^0 \hfill \\](http://latex.codecogs.com/png.latex? = \pi \left[ {\frac{{x^5 }}<br />
{5} + x^4 - 3x^3 - 18x^2 } \right]_{ - 3}^0 \hfill \\)
sign error in this next step ... +162 ?![= \pi \left[ {\left( {\frac{{\left( { - 3} \right)^5 }}<br />
{5} + \left( { - 3} \right)^4 - 3\left( { - 3} \right)^3 - 18\left( { - 3} \right)^2 } \right) - \left( {0 + 0 - 0 - 0} \right)} \right] \hfill \\](http://latex.codecogs.com/png.latex? = \pi \left[ {\left( {\frac{{\left( { - 3} \right)^5 }}<br />
{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}}<br />
{5} + 81 + 81 + 162} \right] \hfill \\](http://latex.codecogs.com/png.latex? = \pi \left[ {\frac{{ - 243}}<br />
{5} + 81 + 81 + 162} \right] \hfill \\)
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