Help with setting up Integrals for volumes

[adelaide87]

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

 

[Gregg]

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

 

[adelaide87]

are a and b correct though?

 

[Theorem.]

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.

 

[Theorem.]

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