# Using disk or ring method, how do you find the volume of y=x^(2)-x, y=3-x^(2), about y=4?

Aug 25, 2015

Integrals to calculate volume have been formed. Integration is left to the student.

#### Explanation:

The sketch showing the region enclosed by the three curves is shown in the figure below The points of intersections of the parabolas would be at x= -1 and x=$\frac{3}{2}$.
The points of intersection of y= ${x}^{2} - x$ and y=4 would be $\frac{1 - \sqrt{17}}{2} \mathmr{and} \frac{1 + \sqrt{17}}{2}$.To calculate the volume of the solid generated by rotating the region about y=4, integration can be done in three segments $\frac{1 - \sqrt{17}}{2} \to - 1 , - 1 \to + \frac{3}{2} \mathmr{and} \frac{3}{2} \to \frac{1 + \sqrt{17}}{2}$

=${\int}_{\frac{1}{2} - \frac{\sqrt{17}}{2}}^{-} 1 \pi {\left(4 - {x}^{2} + x\right)}^{2} \mathrm{dx}$

+${\int}_{-} {1}^{\frac{3}{2}} \pi {\left(4 - 3 + {x}^{2}\right)}^{2} \mathrm{dx}$+${\int}_{\frac{3}{2}}^{\frac{1}{2} + \frac{\sqrt{17}}{2}} \pi {\left(4 - {x}^{2} + x\right)}^{2} \mathrm{dx}$