| Anish
Patel |
hi, i was just looking through one of dr. siklos' booklets and for one of the questions he seems to quote the result 'integral of 2*pi*y*x dx' for the volume formed by rotating a curve about the y axis,i've never seen this-is it something i should know?i think we got taught a formula in P2, something like 'integral of pi*f(y) squared dy',any difference? Any clarification would be much appreciated,thanks! Anish |
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| Ian
Short |
Was the curve y=x2 ? |
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| Alice
Thompson |
I think we're looking at question 22: The function f satisfies the condition f¢(x) > 0 for a £ x £ b, and g is the inverse of f. By making a suitable change of variable, prove that òab f(x) dx=bb-aa-òab g(y)dy Where a = f(a) and b = f(b). Interpret this result geometrically, by means of a sketch, in the case where a and a are both positive. Prove similarly and interpret the formula 2pòab x f(x) dx=p(b2b)(a2a) -pòab[g(y)]2 dy |
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| Ian
Short |
You won't have to know this formula off by heart, only understand the geometric significance of it. Have you answered all parts of the question except this? I'm guessing that you have got to the final 'interpret the formula' bit and have drawn a suitable graph. Consider the area below the graph, above the x-axis and between y=a and y=b. Imagine rotating this about the y-axis. You have an annular base and varying height. In the formula òab 2px f(x) dx, - f(x) is the height - dx is width of infinitesimally small annulus at x - 2px is the circumference of this infinitesimally small annulus. Multiply them for a small slice of volume and the integral sums these volumes. That's the gist of things. Ian |
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| Anish
Patel |
Ah,makes sense,cheers Ian |