Find the volume of the solid that results when the region to the left of the y-axis and inside the ellipse+(y^2/b^2)=1)
a is not equal to b
It is revolved around the y-axis.
I don't know where to start or what equations to use or what to find. I'm further stumped due to the lack of numbers. Please help. Also, how would the problem change if it was revolved around the x-axis or to the right of the y-axis.
Thank you,
Derek
This ellipse has the coordinate axes as its major and minor axes, so it is "centered" at the origin. Think of the solid it forms as an American football or as a rugbny ball. If you were to rotate it around the y-axis - regardless of whether you rotate the portion to the left or right (due to symmetry) - it would look like the top half of the ball. And if you were to rotate it around the x-axis - again regardless of whether you rotate the top portion or the bottom protion (dur to symmetry) you would get the right or left half of the ball. Of course, these could be reversed depending on the values of a and b in the equation, but that's the idea.Originally Posted by DerekZ10
Find the volume of the solid that results when the region to the left of the y-axis and inside the ellipse
a is not equal to b
It is revolved around the y-axis.
I don't know where to start or what equations to use or what to find. I'm further stumped due to the lack of numbers. Please help. Also, how would the problem change if it was revolved around the x-axis or to the right of the y-axis.
Thank you,
Derek
To answer your question, assume you take a small slice of the ellipse in the second quadrant, where. This is the 'width' of the slice from the y-axis. The height of the slice is
. Since a and b are constants, you would treat them just as if they were 2 and 3.
So take the definite integral of(from y=0 to y=b defines the entire part of the ellipse in the 2nd quadrant). You may want to review trigonometric substitution to get this integral!
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