# Homework Help: Help with setting up Integrals for volumes

1. Jun 27, 2010

1. The problem statement, all variables and given/known data

Hi

I am trying to find the volume of a shape.I dont need help to solve, but I would like a hand setting these integrals up. Ive 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

3. 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

2. Jun 27, 2010

### Gregg

Since the equation for the usual revolution is about x or y = 0 you can just do

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

$$\pi \int^b_a (y-1)^2 dx$$

3. Jun 27, 2010

are a and b correct though?

4. Jun 30, 2010

### Theorem.

Well first consider the area bounded by the 3 expressions:

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

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

(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) isnt 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: May 4, 2017
5. Jun 30, 2010

### 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:
$$\pi \int\limits_0^1\ [(2-y^2)^2-(y)^2] dy$$