Help with setting up Integrals for volumes

  • Thread starter Thread starter adelaide87
  • Start date Start date
  • Tags Tags
    Integrals Volumes
adelaide87
Messages
24
Reaction score
0

Homework Statement



Hi

I am trying to find the volume of a shape.I don't need help to solve, but I would like a hand setting these integrals up. I've only recently started doing volumes, so bear with me.

Using the curves:
y=x
x=2-y^2
y=0

Indicate the method used and set the integrals up (dont have to integrate) that give the volume of the solid by rotating the region around:

a) the x axis
b) the y axis
c) the line x=-2
d) the line y=1


The Attempt at a Solution



a) Use the DISK method, Integrate from 0 > 2

(pi)(y^2) dx

= (pi)(x-sqrt(2-x))^2 dx

b) Use SHELL method from 0 > 2

2(pi)(x)(y)

= 2(pi)(x)(x-sqrt(2-x))

c) and d)

Not totally sure, would I just have to incorporate that line into the formula? That is confusing me.

If someone could let me know if a and b are right, and give some guidance in c and d that would be great.

Thanks
 
Physics news on Phys.org
Since the equation for the usual revolution is about x or y = 0 you can just do


\pi \int^b_a (x+2)^2 dy

\pi \int^b_a (y-1)^2 dx
 
are a and b correct though?
 
Well first consider the area bounded by the 3 expressions:

[PLAIN]http://j.imagehost.org/t/0558/math.jpg

y=x, x=2-y^2, y=0

(a) For part a, although you'd normally be able to do a simple disk method, notice that the height of the area bounded by the curves (the radius of rotation) isn't always x=2-y^2, it is also y=x at times. For this reason, your solution to part a is not entirely correct. If you choose to still do simple revolution disk method, you have to split up the integral into two parts, finding where y=x and x=2-y^2 intersect. An alternative method is to use cylindrical shells. and write your integral in terms of y.

1. Find points of intersection (in order to write your limits).
x=2-y^2, x=y (set equal) y=2-y^2 implies y^2+y-2=0 from this simple quadratic we can see that the point of intersection will be (1,1).
As we are using cylindrical shells and rotating about the x-axis, our new limits are :
upper=1
lower=0
2. our integral is now quite easy to set up:
2\pi \int\limits_a^b\ y [f(y)-g(y)] dy
2\pi \int\limits_0^1\ y [(2-y^2)-(y)] dy

(b) For part b you made the same error as in part a, using cylindrical shells in this case means splitting up the integral, which you hadn't accounted for. Use the washer method.
 
Last edited by a moderator:
Using the washer method, the area bounded by y=x, x=2-y^2, y=0 rotated about the y-axis generated a volume which can be represented by the following integral:
\pi \int\limits_0^1\ [(2-y^2)^2-(y)^2] dy
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top