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March 27, 2015

March 27, 2015

Posted by **Jillian** on Saturday, April 5, 2008 at 5:59pm.

1. Find the volume of the solid formed by rotating around the x-axis the region enclosed by the graphs of y = 1 + SQRT(x), the x-axis, the y-axis, and the line x = 4.

a. 7.667

b. 9.333

c. 22.667

d. 37.699

e. 71.209

2. Find the volume of the solid formed by rotating around the y-axis the region bounded by y = 1 + SQRT(x), the y-axis, and the line y = 3.

a. 6.40

b. 8.378

c. 20.106

d. 100.531

e. 145.77

3. Find the volume of the solid formed by rotating around the line y = 5 the region bounded by y = 1 = SQRT(x), the y-axis, and the line y = 3.

a. 13.333

b. 17.657

c. 41.888

d. 92.153

e. 242.95

4. The base of a solid is the region enclosed by the graph of x^2 + 4y^2 = 4 and cross-sections perpendicular to the x-axis are squares. Find the volume of this solid.

a. 8/3

b. 8 pi/3

c. 16/3

d. 32/3

e. 32 pi/3

5. Find the volume of the solid formed by rotating the graph x^2 + 4y^2 = 4 about the x-axis.

a. 8/3

b. 8 pi/3

c. 16/3

d. 32/3

e. 32 pi/3

- Calculus -
**Reiny**, Saturday, April 5, 2008 at 6:52pmI will do the first two for you

1. Vol = pi(integral) y^2 by dx from 0 to 4

= pi (integral) (1 + 2x^1/2 + x)dx from 0 to 4

= pi[x + (4/3)x^3/2 + (1/2)x] from 0 to 4

= pi[4 + 32/3 + 8 - 0]

=71.209

2. from your y = 1 + √x you will need x^2 since you are rotating about the y=axis

y-1 = √x

(y-1)^4 = x^2

vol = pi (integral) (y-1)^4 dy from 1 to 3

= pi[1/5(y-1)^5] from 1 to 3

= pi/5( 32 - 0]

= 20.106

- Calculus -
**Reiny**, Saturday, April 5, 2008 at 7:03pmHow about the last one?

#5.

You have an ellipse rotated about the x-axis

the vertices are (-2,0) and (2,0)

so because of the symmetry I will find the volume from x=0 to x=2 and double it.

from the equation x^2 = -x^2 /4 + 1

so volume = 2pi(integral)((-x^2)/4 + x)dx from 0 to 2

= 2pi[(-1/12)x^3 + x] from 0 to 2

= 2pi[ -8/12 + 2 - 0]

= (8/3)pi

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