Volume of revolution problem
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Volume of revolution problem

[From: ] [author: ] [Date: 13-04-02] [Hit: ]
-Yes that is a cone.You have the right idea but the wrong line.It is an equilateral triangle and on the x-axis, the base is length 2a.Therefore sides are length 2a.If you draw a line down the middle of the triangle,......
The base of a certain solid is the circle x^2+y^2=a^2. Each plane section of the solid cut out by a plane perpendicular to the y-axis is an equilateral triangle with one leg on the base of the sold. Find the volume of the solid.

I wasn't sure where to start. I tried to visualize this shape and the only thing I can come up with is possibly the shape that is being formed is a CONE. If so, that would mean I could get this shape by drawing a line from a height of (0, y) to (y/2, 0) and revolving that around the y-axis?

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Yes that is a cone.

You have the right idea but the wrong line. It is an equilateral triangle and on the x-axis, the base is length 2a. Therefore sides are length 2a. If you draw a line down the middle of the triangle, you can calculate the height. By the Pythagorean theorem, height is sqrt((2a)² - a²) = a√3. So the line you are interested in connects the points (0, a√3) and (a, 0). Now revolve that around the y-axis.

The line would be y = (a√3-0)/(0 - a) (x -a) = (√3/a) (a - x).

There are different ways to set this integral up. But one way is ∫_[0,a√3] πx² dy. Have to substitute that x² with a function in terms of y.
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