I'm in Mu Alpha Theta for my school.. and got this question on the MAO test that we got today. Now.. I'm used to revolution around vertical lines and horizontal lines, as well as axii.. but how does one revolve around another function? Specifically, a slanted line?
The exact question:
What is the volume of the region formed when the ellipse (9x^2+4y^2=36) is revolved around the line (2x+y=5)?
The exact question:
What is the volume of the region formed when the ellipse (9x^2+4y^2=36) is revolved around the line (2x+y=5)?
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The first thing I'd think of would be to rotate the whole plane so that the line is vertical or horizontal, but that would make the region rather ugly to integrate. The second idea I had is to use the Theorem of Pappus, which states that the volume of a solid of revolution is equal to the area of the planar figure being rotated times the distance traveled by its geometric centroid (where this distance is measured along a perpendicular line to the axis of revolution. This process would make this whole question notoriously easy.
We know that the center of the ellipse it's at (0,0), and a perpendicular line to y = 5-2x through (0,0) is y = x/2
These two lines intersect at:
x/2 = 5 - 2x
x = 10 - 4x
x = 2, y = 1
The distance between the origin and the point (2,1) is √5. This is the radius of the circle traveled by the center of the ellipse, and so the whole distance traveled is 2π√5. To find the area of the ellipse we'll need to know the lengths of the major and minor axes, so manipulate the equation to look like
(x²/4) + (y²/9) = 1
and now that it's in standard form we have that the minor axis (along the x-axis) is of length 2, and the major (along the y-axis) is of length 3. Therefore the area of this ellipse is π(2)(3) = 6π, and so by the Theorem of Pappus we have that the volume of revolution is
6π * 2π√5 = 12π²√5
In general, the "washer" or "ring" method could be applied as before, but as we saw the radius has to be calculated with a perpendicular, so the integrand would get very complicated very quickly.
We know that the center of the ellipse it's at (0,0), and a perpendicular line to y = 5-2x through (0,0) is y = x/2
These two lines intersect at:
x/2 = 5 - 2x
x = 10 - 4x
x = 2, y = 1
The distance between the origin and the point (2,1) is √5. This is the radius of the circle traveled by the center of the ellipse, and so the whole distance traveled is 2π√5. To find the area of the ellipse we'll need to know the lengths of the major and minor axes, so manipulate the equation to look like
(x²/4) + (y²/9) = 1
and now that it's in standard form we have that the minor axis (along the x-axis) is of length 2, and the major (along the y-axis) is of length 3. Therefore the area of this ellipse is π(2)(3) = 6π, and so by the Theorem of Pappus we have that the volume of revolution is
6π * 2π√5 = 12π²√5
In general, the "washer" or "ring" method could be applied as before, but as we saw the radius has to be calculated with a perpendicular, so the integrand would get very complicated very quickly.
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Here's a hint
Think about how to calculate the volume of 9x^2+4y^2=36 rotated about the x axis
To rotate about a different axis, 2x+y=5, change the coordinates so that the x axis is in the direction of the line.
Coordinates are given by [xnew; ynew] = [cos theta, sin theta; -sin theta cos theta][xorig yorig]
You can get theta from the slope, but need to subtract off the x intercept in order to rotate about the new origin.
It would probably help to draw it out on graph paper.
Think about how to calculate the volume of 9x^2+4y^2=36 rotated about the x axis
To rotate about a different axis, 2x+y=5, change the coordinates so that the x axis is in the direction of the line.
Coordinates are given by [xnew; ynew] = [cos theta, sin theta; -sin theta cos theta][xorig yorig]
You can get theta from the slope, but need to subtract off the x intercept in order to rotate about the new origin.
It would probably help to draw it out on graph paper.