Calculus Help!!

posted by .

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.

  • Calculus Help!! -

    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

  • Calculus Help!! -

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

  • Calculus Help!! -

    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.

  • Calculus Help!! -

    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.

  • Calculus Help!! -

    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)

  • Calculus Help!! -

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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. calculus

    let R be the region bounded by the graphs of y = sin(pie times x) and y = x^3 - 4. a) find the area of R b) the horizontal line y = -2 splits the region R into parts. write but do not evaluate an integral expression for the area of …
  2. calculus

    The figure shows the region bounded by the x-axis and the graph of . Use Formulas (42) and (43)-which are derived by integration by parts-to find (a) the area of this region; (b) the volume obtained by revolving this region around …
  3. Calculus

    1. Find the area of the region bounded by the curves and lines y=e^x sin e^x, x=0, y=0, and the curve's first positive intersection with the x-axis. 2. The area under the curve of y=1/x from x=a to x=5 is approximately 0.916 where …
  4. Calculus AP

    Let R be the region bounded by the graphs of y=cos((pi x)/2) and y=x^2-(26/5)x+1. A. Find the area of R. B. The vertical line x=k splits the region R into two equal parts. Write, but do not solve, an equation involving integrals that …
  5. calculus

    3). The shaded region is bounded by the y-axis and the graphs of y=1+√x, y=2. Find the volume of the solid obtained by rotating this region around the x-axis. Answer choices: 7/6pi, 4/3pi, 11/6pi, 5/3pi, 13/6pi, 5/6pi 4). Find …
  6. Calculus

    We're learning disks, shells, and cylinders in school but we have a substitute and I've been trying to teach this to myself. Can you check them please?
  7. Calculus

    We're learning disks, shells, and cylinders in school but we have a substitute and I've been trying to teach this to myself. Can you check them please?
  8. CALCULUS HELP PLZ

    The region R is bounded by the x-axis, y-axis, x = 3 and y = 1/(sqrt(x+1)) A. Find the area of region R. B. Find the volume of the solid formed when the region R is revolved about the x-axis. C. The solid formed in part B is divided …
  9. CALCULUS

    The region R is bounded by the x-axis, y-axis, x = 3 and y = 1/(sqrt(x+1)) A. Find the area of region R. B. Find the volume of the solid formed when the region R is revolved about the x-axis. C. The solid formed in part B is divided …
  10. Calculus

    'Find the volume of the solid generated by revolving the region bounded by x=16-y^2, x-axis, and y-axis around the x-axis.' I know in theory how to do it, but I'm a little confused by what constitutes the bounded region because it's …

More Similar Questions