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# 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 – x^{2 }and y = e^{x} 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 – x^{2} is the function at the top so let it be “R”

y = e^{x} is the function at the bottom so let it be “r”

1.0580

# Area R = S [(4 – x^{2}) – (e^{x})] dx

0

1.0580

**Area R**** =** [4x - 1/3x^{3} – e^{x}]

0

Now you antidifferentiate.

1.0580

**Area R**** =** [4x - 1/3x^{3} – e^{x}]

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 units^{2}

**b) When R is revolving about the x-axis, use the **__washer method__ to find the volume.

**V =** π S (R^{2}-r^{2}) dx

1.0580

= π S ((4 – x^{2})^{ 2}-(e^{x})^{2})^{ }dx

0

1.0580

= π S (16 - x^{4} – e^{2x}) dx

0

1.0580

** =** π [16x – 1/5x^{5} – e^{2x}/2]

0

** =** π ([16(A) – 1/5(A)^{5} – e^{2(A)}/2] - [16(0) – 1/5(0)^{5} – e^{2(0)}/2])

** =** π ([12.5140] - [0 – 0 – 1/2])

** =** *40.8846 units*^{3}

**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 – x^{2 }+1) - (e^{x }+ 1)) dx

0

1.0580

= 2π S x ((5 - x^{2}) – (e^{x} + 1)) dx

0

1.0580

= 2π S ((5x - x^{3}) – (xe^{x} + x)) dx

0

1.0580

= 2π S (4x – x^{3} - xe^{x}) dx

0

1.0580

** =** 2π [2x^{2} – 1/4x^{4} - (xe^{x -}e^{x})]

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 units*^{3}

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