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)3 – e(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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