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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 xaxis when y=square root of x, y=1 and x=4
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 xaxis when y=square root of x, y=1 and x=4

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think it turns into a disk but I am not sure. And I cant figure out the integral.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328252349257:dw this uses the washer method where for each circular crosssection you take large area and subtract smaller area Here the outer radius is sqrt(x), the inner radius is 1 \[V = \pi \int\limits_{?}^{?}R^{2} r^{2} \] \[V = \pi \int\limits_{1}^{4}(\sqrt{x}^{2}  1^{2}) dx = \pi \int\limits_{1}^{4}(x  1)dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whatever the axis you're revolving it around, you just need to determine the radius or the distance from the axis to the curve

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0example: 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so if I revolved it around the x=2 my integral would be pi ((sqrt(x)2)^2(12)^2)dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry typo there...it would be y=2, in other words a horizontal axis no add 2 to the radius, not subtract 2 \[\pi \int\limits_{?}^{?} (\sqrt{x}+2)^{2}  3^{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328253465405:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but I thought since x=2 thats why you would really subtract 2 from the square root and the 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no x does not equal 2 instead of revolving around xaxis, 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay that makes more sense

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you for all of your help
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