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Homework Help: A volume problem

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    f(x)=kx2+x3, where is k is a positive constant. Let R be the region in the first quadrant bounded by the graph of F and the x-axis.

    a. Find all values of the constant k for which the area of R equals 2.
    b. For k>0, write an integral expression in terms of k for the volume of the solid generated when R is rotated around x-axis.
    c. For k>0, write an expression in terms of k, involving one or more integrals, that gives the perimeter of R.

    2. Relevant equations
    V= pi *r2


    3. The attempt at a solution

    a. For a. I find the interception of F(x) and the x-axis. After that, I set up an integral of f(x) evaluating from 0 to k and set that integral equal to 2. Am I right.

    b. and c. I don't really get the part of in terms of k. so I just do a normal integration of dish method of f(x) and I substitute the number in part a for k. Am I right ? Or is it trickier ?
     
  2. jcsd
  3. Nov 8, 2008 #2

    Mark44

    Staff: Mentor

    Are you sure you have the right formula for f? The formula that you gave, f(x) = kx^2 + x^3, has an intercept at (0, 0) and no others on the positive x-axis. This means that the area in the first quadrant between the graph of f and the x-axis is infinite.

    In you attempt to find the answer to part a, why would you have k as a limit of integration? Before going on to parts b and c, make sure that you are giving us the correct information for this problem.
     
  4. Nov 8, 2008 #3
    I am sorry it's minus not plus.

    Because k is a x-intercept ?

    Am I right ?
     
  5. Nov 8, 2008 #4

    HallsofIvy

    User Avatar
    Science Advisor

    What's "minus not plus"!

    Do you mean it should be y= kx2- x3? In that case y= x2(k- x) so x= 0 is a double root and x= k is a root.

    If you rotate the curve, for 0< x< k, around the x-axis each cross section perpendicular to the x-axis is a circle with radius y= kx2- x3. The area of such a circle is [itex]k\pi y^2[/itex].

    The perimeter of R consists of two parts: the curve between x= 0 and x= k and the straight line from (0,0) to (k, 0).
     
  6. Nov 8, 2008 #5
    Can you explain about part b ? Why there is k there ? I got only pi*^2
     
  7. Nov 9, 2008 #6

    Mark44

    Staff: Mentor

    In part b, the volume of a typical volume element is
    [tex] \Delta V = \pi y^2 \Delta x[/tex]
    and not [tex]k \pi y^2 [/tex] as Halls said.

    or,
    [tex] \Delta V = \pi (x^2(k - x))^2 \Delta x[/tex]

    So what should the integral look like? (All you need to do is write the integral.)

    I took the problem further and calculated the integral, getting:
    [tex] \pi k^7 [\frac{1}{5} - \frac{1}{3} + \frac{1}{7}][/tex]
     
  8. Nov 9, 2008 #7

    HallsofIvy

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    Science Advisor

    I have no idea where the k came from! It snuck in while I wasn't looking!
     
  9. Nov 9, 2008 #8
    what do you guys think the " in term of k" means ? Do I just write a normal integral or is it trickier ?
     
  10. Nov 9, 2008 #9

    Mark44

    Staff: Mentor

    "In terms of k" means that when you write the integral expressions, they will involve k, which is unspecified. For parts b and c, you don't actually have to evaluate the integrals--at least that's how I interpret the instructions.
     
  11. Nov 9, 2008 #10
    So do I substitute the k I found in part a in or just leave it as k ?
     
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