Calculus

Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = −3.

  1. 👍 0
  2. 👎 0
  3. 👁 420
asked by Michael
  1. The region is a triangular patch with vertices at (1,0), (1,1), and (1,e)
    So the volume is a stack of washers of thickness dx:
    v = ∫[1,e] π(R^2-r^2) dx
    where R=(1+3) and r=(lnx+3)
    v = ∫[1,e] π(4^2-(lnx+3)^2) dx

    Or, you can set it up as a stack of nested shells of thickness dy
    v = ∫[0,1] 2πrh dy
    where r=y+3 and h=x-1=e^y-1
    v = ∫[0,1] 2π(y+3)(e^y-1) dy

    1. 👍 0
    2. 👎 0
    posted by oobleck

Respond to this Question

First Name

Your Response

Similar Questions

  1. calculus review please help!

    1) Find the area of the region bounded by the curves y=arcsin (x/4), y = 0, and x = 4 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative. 2)Set up, but do not evaluate,

    asked by Danni on March 22, 2017
  2. calculus

    1. Find the area of the region bounded by f(x)=x^2 +6x+9 and g(x)=5(x+3). Show the integral used, the limits of integration and how to evaluate the integral. 2. Find the area of the region bounded by x=y^2+6, x=0 , y=-6, and y=7.

    asked by katarina on May 15, 2012
  3. calculus

    using the method of shells, set up, but don't evaluate the integral, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. Y=e^x, x=0, y=2, about y=1

    asked by marks on March 16, 2014
  4. calc 3

    1. Evaluate the given integral by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 36x^2+25y^2=1. L= double integral R (4sin(144x^2+100y^2) dA. 2. Use the given

    asked by lala on November 4, 2014
  5. Calculus

    Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -3 I got the integral from 1 to 2.718 of pi(1)^2-pi(ln(x))^2 Is

    asked by Kait on March 19, 2018
  6. Calculus check

    The functions f and g are given by f(x)=sqrt(x^3) and g(x)=16-2x. Let R be the region bounded by the x-axis and the graphs of f and g. A. Find the area of R. B. The region R from x=0 to x=4 is rotated about the line x=4. Write,

    asked by Anonymous on April 16, 2015
  7. Calculus

    Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 2, and x = 1 is revolved around the line y = −2.

    asked by Jared on May 18, 2017
  8. calculus

    Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = −3.

    asked by Anonymous on April 16, 2015
  9. Calculus

    Use the shell method to set up, but do not evaluate, an integral representing the volume of the solid generated by revolving the region bounded by the graphs of y=x^2 and y=4x-x^2 about the line x=6. I had the shell radius as

    asked by Ryan on March 7, 2011
  10. Calculus Please Check my answer

    Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -1. ANSWER: v = ∫[1,e] π(4-(lnx+1)^2) dx

    asked by Dloc on May 2, 2017

More Similar Questions