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Question 7

Page history last edited by PBworks 16 years, 10 months ago

 

Question 7

 

You MAY use a calculator to solve this problem.

 

 

 

 

Let R be the region in the first quadrant shown in the figure.

 

 

(a) Find the area of R.

 

 

(b) Find the volume of the solid generated when R is revolved about the x-axis.

 

 

(c) Find the volume of the solid generated when R is revolved about the line x = -1.

 

 

 

 

 

 

 

 

Solution

 

a) To find the area between these two functions:

 

1. Graph the functions into you graphing calculator

 

 

 

2. Find the point of intersection.  To do this, go to [2nd] [trace] 5:intersect.

 

We find that the point where y = 4 – x2 and y = ex intersect is (1.0580064, 2.880622). Store the x value into [ALPHA] A.

Rounded off to four decimal places the point of intersection is (1.0580, 2.8806).

 

3.  Write your integral for area.  You know that the area is bounded by the y-axis and the point of intersection since it is in the first quadrant.  Therefore, your lower limit is 0 and your upper limit is 1.0580.

 

      b

Area R = S (R-r) dx

      a

 

 

y = 4 – x2 is the function at the top so let it be “R

y = ex is the function at the bottom so let it be “r 

 

 

        1.0580

Area R = S [(4 – x2) – (ex)] dx

              0 

 

 

                                            1.0580

Area R = [4x - 1/3x3 – ex]

                             0

 

 

Now you antidifferentiate.

 

 

                                           1.0580

Area R = [4x - 1/3x3 – ex]

                             0

 

Note: When you plug in the values, use [ALPHA] A for a more accurate answer.

 

Area R = [4(A) – 1/3(A)3 – e(A)] – [4(0) – 1/3(0)3e(0)]

 

            = [0.9566] - [-1]

 

            = 1.9566 units2

 

 

 

b) When R is revolving about the x-axis, use the washer method to find the volume.

 

 

 

 

V = π S (R2-r2) dx

           

              1.0580

   = π S ((4 – x2) 2-(ex)2) dx

             0

 

             1.0580

   = π S (16 - x4 – e2x) dx

              0

                                                  1.0580

   = π [16x – 1/5x5 – e2x/2]

                                                 0

 

 

   = π ([16(A) – 1/5(A)5 – e2(A)/2] - [16(0) – 1/5(0)5 – e2(0)/2])

 

   = π ([12.5140] - [0 – 0 – 1/2])

 

   = 40.8846 units3

 

 

c)  When R is revolving around x=-1, use the shell method to find the volume.

 

 

 

 

 

V = 2π S x ((R+1)-(r+1)) dx

            

               1.0580  

   = 2π S x ((4 – x2 +1) - (ex + 1)) dx

                0

             

                1.0580                      

    = 2π S x ((5 - x2) – (ex + 1)) dx

                0

 

              1.0580

   = 2π S ((5x - x3) – (xex + x)) dx

                0

 

              1.0580

   = 2π S (4x – x3 - xex) dx

                0

 

                                                     1.0580

 = 2π [2x2 – 1/4x4 - (xex -ex)]

                                   0

 

 = 2π ([2(A)2 – 1/4(A)4 - ((A)e(A) – e(A))] - [2(0)2 – 1/4(0)4 - ((0)e(0) – e(0))]

 

 

 = 2π ([2(A)2 – 1/4(A)4 - ((A)e(A) – e(A))] - [0 – 0 - (0-1)])

 

 

= 2π ([1.7584]- [1])

 

= 2π [0.7584]

 

= 4.7652 units3

 

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