Calculus Help!!

Region R is bounded by the functions f(x) = 2(x-4) + pi, g(x) = cos^-1(x/2 - 3), and the x axis.

a. What is the area of the region R?

b. Find the volume of the solid generated when region R is rotated about the x axis.

c. Find all values c for f(x) and g(x) in the closed interval p <= c <= q for which each function equals the average value in the indicated interval.

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  1. the curves intersect at (4,π)

    a = ∫[0,4] 2(x-4)+π dx
    + ∫[4,8] arccos(x/2-3) dx
    = x^2+(π-8)x [0,4]
    + x arccos(x/2-3) + 6 arcsin(x/2-3) - √(4-(x-6)^2)
    = 4π-16 + 2π
    = 6π-16

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  2. Ok, that makes sense. How would I find the answers to b and c though? I'm not even sure where to start...

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  3. you must have some idea where to start.
    see wolframalpha.com for doing integrals

    v = ∫π y^2 dx for each region.
    = π∫[0,4](2(x-4)+π)^2 dx + π∫[4,8](arccos(x/2-3))^2 dx

    for the avg value, not sure what p and q are supposed to be.

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  4. P and Q are the x values on the graph I was given for this problem where the two functions intersect the x axis. They don't have any numerical assignments.

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  5. the average value over [0,8] is k=(6π-16)/8

    There are two x values where this occurs

    f(x) = k and g(x) = k

    to see where these are visit wolframalpha.com and enter

    plot y=2(x-4)+pi , y=arccos(x/2-3), y=(6*pi-16)/8, x=0..8

    f(c) = (6π-16)/8 = 2(c-4)+π
    c = 3 - π/8

    g(c) = (6π-16)/8 = arccos(c/2-3)
    c = 6+2cos(3π/4 - 2)

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  6. Newsflash: 2(x-4)+ pi doesn't cross the x-axis before 0 so you can't use zero for the integral

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