Computing the volume of a solid shape that is created when you rotate a region of a curve 360 degrees around an axis is a problem presented to calculus students during the process of learning integrals. This problem is solved by adding the dimension of rotation using the formula of circular area (pi * radius squared), where the radius in this case is the curve's function to the normal integral formula. This rotational volume equation becomes the definite integral function of: volume = the integral between point a and b of pi * f(x) squared dx (with respect to x).
Instructions
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1
Use the curve described by the equation y = x^2 in this example and find the volume of the rotational solid bounded by the points x = 0 and x = 2.
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2
Set up the integral equation as volume = integral between x = 0,2 of pi * (x^2)^2 dx. This becomes integral of pi * x^4 dx.
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3
Integrate the function, which becomes pi * (1/5)(x^5).
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4
Insert our boundary coordinates x = 0,2 and compute the area. This becomes: (pi*(2^5)/5) - 0. The result is then 32pi/5.