I will be the first to admit that I'm totally feeling lost right now in
my Calculus class. Integrands have begun to haunt my dreams. That aside,
I've done one of my assigned problems, and I';m almost positive I'm so lost
that my answers are no where near what they should be.
The problem has 4 parts, and the question reads:
For the region given by the shaded area in the figure below:
[Here is a figure of a trapezoidal shape going from (1,1) to (2,4) with the
"top" line along the hyperbola y = x^2]
A. Find the area of the shaded region.
I used 1/2h(b1+b2) to find f(x), and h=1, b1=1, b2=4, I came up with:
1/2(1)(1+4) = 1/2(5) = 5/2
Then, finding the integrand for 5/2 x^2 dx:
5/2[(x^3)/3] and evaluated from 1 to 2 for:
5/2(8/3 - 1/3) = 5/2(7/3) = 35/6
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
B. Find the volume of the solid generated by rotating about the x axis.
The graph I found for this resembles the bell of a trumpet beginning
at x = 1 going to x = 2,
at x=1 y goes from -1 to 1
at x=2 y goes from -4 to 4
This is what I've done, but I'm not sure if it is correct:
Using V = (pi)r^2h, and h = 1, I come up with (pi)(x^2)^2(h) = (pi)x^4
Then integrating from 1 to 2 I get:
(pi) * the Integrand x^4dx
=(pi)[(1/5 x^5)] evaluated at 1 & 2 for:
(pi)[32/5 - 1/5] = (pi)31/5
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
C. Find the volume of the solid generated by rotating the region about the
y axis.
The graph I found for this resembles a cylinder with an inner radius of
(-1,1) to (1,1) and an outer radius of (-2,4) to (2,4) and of course
hollow in the center from x = -1 to x = 1.
Again, not sure if I'm on the right track with this:
Finding the integrand for 2(pi)x(x^2)dx
=2(pi)[1/4 x^4] and evaluated from 1 to 2 for:
2(pi)[16/4 - 1/4] = 2(pi)[15/4] = [5(pi)]/2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
D. Find the volume of the solid generated by rotating the region about the
line x = 2
using V = (pi)r^2h
for y = x^2, with A(y) = (pi)(sq.root of y) , and y = 0 to y = 4
Finding the integrand for (pi)(sq.root of y)dy [(pi)*y^(1/2)dy]
= (pi)[2/3y^(3/2) and evaluating at 0 and 4 for:
(pi){[2/3(4)^(3/2)] - [2/3(0)^(3/2)]}
= [16(pi)]/3
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
After your laughter has subsided, and you've wiped the tears from your eyes, PLEASE let me know where I've gone wrong.
If you're really bored, and just dying to do more, let me know and I'll e-mail you the other problems I've got questions about!