Qs 1 of the 2008 AP Calculus AB exam,
found here
http://www.google.ca/search?q=AP+Cal...ient=firefox-a
(1st link)
why is a volume of the solid, the integral of the height of the function squared?
(part C)
why is the volume of the pond, the integral of the height * depth?
(part D)
im not understanding the theory.
When we do volumes of solids of revolution, we're basically taking cross-sections of circles.
Draw the line y = 2 - x. Revolve that line 360 degrees around the x-axis and you'll get a cone. Now cut that cone into infinitely small sections. What you're left with is an infinite number of circles (two-dimensional, because we've split the cone infinitely).
If you examine each of those circles, you'll notice that they have a radius equivalent to the value of f(x). Like, at the base of the cone, you'd have a circle with area f(0) or 2 - 0 or 2. And at the tip, f(2) or 0.
So, if the function f(x) represents our radius at any given point for that solid, than if we integrate from 0 to 2, we'll be expanding f(x) over an infinite span of dx. Basically saying, we get infinite circles from 0 to 2 (we meaning we reverse to "cutting"). This, in essence, is no more than adding an extra dimension (the 3rd) to our two-dimensional circle.
Well if the cross section of the solid is a square that means that height=base, and therefore the area of the cross section is just. Well, that's just one cross section, so to find the total volume, you need to find the integral on the given interval. And what's the base? Well, it's just the region R, which can be written as
. So the volume of the solid with a square cross-section is just
-x^{3}+4x)^{2}\,\,dx)
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