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

## Answer this Question

## Related Questions

- Integral calculus - Please can anyone help with the following problems - thanks...
- Calculus - Find the volume V of the solid obtained by rotating the region ...
- Integrating Factors - I've been working on this hw problem for a while now, but ...
- Calculus please help!!! double integral - Combine the following two integrals ...
- calculus - find the volume of solid inside the paraboloid z=9-x^2-y^2, outside ...
- double integrals - Combine the following two integrals into one by sketching the...
- Calculus - Let R be the region bounded by the curve x=9y-y^2 and the y- axis. ...
- Calculus - Volume By Integration - Find the volume of the solid generated by ...
- Calculus - Evaluate the triple integral _E (xy)dV where E is a solid tetrahedron...
- calculus - Find the volume V of the solid in the rst octant bounded by y = 0, z...

More Related Questions