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[edwardraff]

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[Frosty3799]

i think i got a 8% on the test for this stuff...

 

[edwardraff]

that helpful

 

[MereMortal]

There are two solutions that satisfy these conditions. One way to go about solving this is the following:

You know by using the disk method that V=pi*integral(a->b) {[f(x)]^2 dx}. First make the problem linear by letting g(x)=[f(x)]^2.

Now you know that the solution is b^2-a*b, so that g(x) is at most a 1st order polynomial. Take it to be g(x) = j*x + k, where {j, k} are coefficients to be found.

If you then stick this form of g(x) into the volume integral and evaluate at the limits, you get

V = b^2/pi - (a*b)/pi = (j*b^2)/2 - (j*a^2)/2 + k*b - k*a

There is one equation and two unknowns, leading to two solutions. You can get these solutions by matching coefficients of the powers of a & b on the RHS and LHS.

Matching the 1st term on the RHS gives j = 2/pi. Substituting this j leads to k = -a/pi.
The first solution is therefore g(x) = (2*x-a)/pi, or f(x) = (+/-)sqrt[(2*x-a)/pi].

Matching the 2nd term on th RHS gives j = 0. It follows that k = b/pi.
The second solution is then g(x) = b/pi, or f(x) = (+/-)sqrt(b/pi).