kay, well i kinda get the idea of what you're supposed to do, but it's just out of reach. can you explain?
Use the disk method to verify that the volume of a right circular cone is 1/3pi*r^2*h, where r is the radius of the base and h is the height.
the pi in the volume formula comes from the pi at the front of the formula for solid of revolution and i think the equation would be something like x^3*y (where r is x and h is y), but that seems wrong somehow. help?
Use the disk method to verify that the volume of a right circular cone is 1/3pi*r^2*h, where r is the radius of the base and h is the height.
the pi in the volume formula comes from the pi at the front of the formula for solid of revolution and i think the equation would be something like x^3*y (where r is x and h is y), but that seems wrong somehow. help?
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Draw a straight line passing through the origin and going to some arbitrary point (h, r) in the first quadrant (so h and r are both positive). The slope of this line is (r/h). The equation is:
y = (r/h)x
Using the disk method we will rotate that line about the x axis. If we do this over the interval of 0 <=x <= h, then we will end up with a right circular cone of height h and a base radius of r.
The disk method states the volume integral should be set up as follows:
V(x) = π∫[R(x)]^2dx
where R(x) is the function of x that represents the radius of the disk at position x. In our problem R(x) = (r/h)x. Therefore:
V(x) = π∫[(r/h)x]^2dx
V(x) = π∫(r^2/h^2)x^2dx
Since r^2/h^2 are constant, they can be pulled out of the integration, giving us:
V(x) = (πr^2/h^2)∫x^2dx
V(x) = (πr^2/h^2)((1/3)x^3)
V(x) = [(πr^2)/(3h^2)]x^3
We must evaluate the above at our end points of the interval of integration, which are x = h and x = 0. Therefore:
V = V(h) - V(0)
V = [(πr^2)/(3h^2)]*h^3 - [(πr^2)/(3h^2)]*0^3
Since [(πr^2)/(3h^2)]*0^3 is 0, then
V = [(πr^2)/(3h^2)]*h^3 - 0
V = [(πr^2)/3]*h
V = πhr^2/3
V = (1/3)πhr^2
That matches the formula they asked you to verify.
y = (r/h)x
Using the disk method we will rotate that line about the x axis. If we do this over the interval of 0 <=x <= h, then we will end up with a right circular cone of height h and a base radius of r.
The disk method states the volume integral should be set up as follows:
V(x) = π∫[R(x)]^2dx
where R(x) is the function of x that represents the radius of the disk at position x. In our problem R(x) = (r/h)x. Therefore:
V(x) = π∫[(r/h)x]^2dx
V(x) = π∫(r^2/h^2)x^2dx
Since r^2/h^2 are constant, they can be pulled out of the integration, giving us:
V(x) = (πr^2/h^2)∫x^2dx
V(x) = (πr^2/h^2)((1/3)x^3)
V(x) = [(πr^2)/(3h^2)]x^3
We must evaluate the above at our end points of the interval of integration, which are x = h and x = 0. Therefore:
V = V(h) - V(0)
V = [(πr^2)/(3h^2)]*h^3 - [(πr^2)/(3h^2)]*0^3
Since [(πr^2)/(3h^2)]*0^3 is 0, then
V = [(πr^2)/(3h^2)]*h^3 - 0
V = [(πr^2)/3]*h
V = πhr^2/3
V = (1/3)πhr^2
That matches the formula they asked you to verify.