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

This problem set is ridiculously hard. I know how to find the volume of a solid (integrate using the limits of integration), but these questions seem more advanced than usual. Please help and thanks in advance!

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 6:52pm
I 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 7:03pm
How 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

- Calculus -
**dj**, Tuesday, December 15, 2015 at 8:03pm
13242

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