Calculus
posted by Jillian on .
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 xaxis the region enclosed by the graphs of y = 1 + SQRT(x), the xaxis, the yaxis, 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 yaxis the region bounded by y = 1 + SQRT(x), the yaxis, 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 yaxis, 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 crosssections perpendicular to the xaxis 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 xaxis.
a. 8/3
b. 8 pi/3
c. 16/3
d. 32/3
e. 32 pi/3

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
y1 = √x
(y1)^4 = x^2
vol = pi (integral) (y1)^4 dy from 1 to 3
= pi[1/5(y1)^5] from 1 to 3
= pi/5( 32  0]
= 20.106 
How about the last one?
#5.
You have an ellipse rotated about the xaxis
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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