Calculus II/III

posted by .

A. Find the integral of the following function.

Integral of (x√(x+1)) dx.

B. Set up and evaluate the integral of (2√x) for the area of the surface generated by revolving the curve about the x-axis from 4 to 9.


For part B of our question , the surface of revolution is is
integral of 2 pi y * dx
= 4 pi x^(1/2) dx
for x from 4 to 9.
The indefinite integral is
4 pi x^(3/2)/(3/2)
For the definite integral, subtract the value at x = 4 from the value at x = 9.

Try making the subsitution u = x + 1
x = u -1
dx = du
The integral becomes

Integral of (u^3/2) - u^(1/2) du
= (2/5) [u^(5/2) - (2/3) u^(3/2)] du
= (2/5) (x+1)^(5/2) - (2/3)(x+1)^(3/2)]

For the part B of you question, integrate
2 sq

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Integral Calculus

    Find the area of the surface generated by revolving the given curve about the y-axis. 8xy^2=2y^6+1 , 1<=y<=2. thank you so much.. :)
  2. Calc

    Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis. y = (x³/6) + (1/2x), 1≤ x ≤ 2
  3. Calc

    Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the y-axis. y = cube rt. (x) + 2 Thank you so much!!
  4. 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 (6-x) …
  5. calculus

    2. Let R be the region in the first quadrant bounded by the graphs of (x^2/9)+(y^2/81)=1 and 3x+y=9 . a. Set up but do not evaluate an integral representing the area of R. Express the integrand as a function of a single variable. b. …
  6. calculus

    2. Let R be the region in the first quadrant bounded by the graphs of (x^2/9)+(y^2/81)=1 and 3x+y=9 . a. Set up but do not evaluate an integral representing the area of R. Express the integrand as a function of a single variable. b. …
  7. asdf

    Find a definite integral indicating the area of the surface generated by revolving the curve y = 3√3x ; 0 ≤ y ≤ 4 about the x – axis. But do not evaluate the integral.
  8. Integral calculus

    find the area of the surface generated by revolving about x-axis the upper half of the ellipse 4x^2 + 16y^2 = 64
  9. Math - Calculus

    Find the area of the surface of revolution generated by revolving the curve y = 3 sqrt (x), 0 <= x <= 4, about the x-axis. Okay, so I've set up the integral like this: 2pi ∫[0,4] (3 sqrt (x))(sqrt(1+(1/4x)))dx Which is coming …
  10. Math

    Let A denote the portion of the curve y = sqrt(x) that is between the lines x = 1 and x = 4. 1) Set up, don't evaluate, 2 integrals, one in the variable x and one in the variable y, for the length of A. My Work: for x: integral[4,1] …

More Similar Questions