## anonymous 4 years ago how do I set up the integral to find the volume of the solid obtained by revolving the region about the x-axis when y=square root of x, y=1 and x=4

1. anonymous

I think it turns into a disk but I am not sure. And I cant figure out the integral.

2. anonymous

|dw:1328252349257:dw| this uses the washer method where for each circular cross-section you take large area and subtract smaller area Here the outer radius is sqrt(x), the inner radius is 1 $V = \pi \int\limits_{?}^{?}R^{2} -r^{2}$ $V = \pi \int\limits_{1}^{4}(\sqrt{x}^{2} - 1^{2}) dx = \pi \int\limits_{1}^{4}(x - 1)dx$

3. anonymous

thank you! but what if I want to switch it and revolve it about the x axis or another point such as x=-1? how do I know how to change my integrals?

4. anonymous

whatever the axis you're revolving it around, you just need to determine the radius or the distance from the axis to the curve

5. anonymous

example: for this region say we revolve around x=-2: my curve is now 2 units further away from the axis, just add 2 to the radius outer would become sqrt(x) +2 inner would become 3

6. anonymous

oops i meant y=-2

7. anonymous

so if I revolved it around the x=-2 my integral would be pi ((sqrt(x)-2)^2-(1-2)^2)dx

8. anonymous

sorry typo there...it would be y=-2, in other words a horizontal axis no add 2 to the radius, not subtract 2 $\pi \int\limits_{?}^{?} (\sqrt{x}+2)^{2} - 3^{2}$

9. anonymous

|dw:1328253465405:dw|

10. anonymous

but I thought since x=-2 thats why you would really subtract 2 from the square root and the 1

11. anonymous

no x does not equal -2 instead of revolving around x-axis, we are revolving around the line y=-2 the distance the function is away from the line y=-2 is now 2 more than the x_axis

12. anonymous

oh okay that makes more sense

13. anonymous

thank you for all of your help

14. anonymous