help with some calc 2
 

So I'm reviewing for an exam that I have tomorrow, and I came across these two similar problems. I realized I completely forgot how to do them. I know they both have to do with the integral of pi R squared, and something to do with outter radius - inner radius, but I am drawing a blank on how exactly to do them. Probably due to cramming, but somebody help me out with these please ---

1. Find the volume of the solid generated by rotating the region bounded by the curves y=xx and x=yy about the y-axis.

2. Find the volume of the solid generated by rotating the region bounded by the curves y=3x-xx and y=0 about the line x= -1.

Again, just trying to figure out how to do these problems. Since I'm reviewing for an exam, the process is much more important than the raw answer.

Re: help with some calc 2
 

I think you use the shell method formula thing and integrate from one line to the other.... I'm posting from my phone so I have no book or google with me... I'll try and remember to look it up when I get back

Re: help with some calc 2
 

You use shell's method for this problem: V=2*pi*integral(p(x)*h(x))

p(x) is the distance from the axis of revolution
h(x) is the height of the differential of area

I set the integrals up for you below. If you have any questions, let me know.

http://img138.imageshack.us/img138/9...s002jy6.th.gif

Re: help with some calc 2
 

You are trying to use the "washer method", which is just as correct as the shell method. In this method, you are slicing your region up into rectangles and revolving the rectangles around your line of revolution.

In the first example, you're revolving around the y-axis, so you'll make your slices perpendicular to the y-axis, i.e. you'll make rectangles of width "dy". (If you want to be nitty, they're width \delta y, which we'll let go to 0).

Now, when you revolve each rectangle around the y-axis, you'll get what looks like a donut - a cylinder with another cylinder removed. It's volume is

(volume of the outer cylinder) - (volume of inner cylinder)

= [pi*(outer radius)^2 - pi*(inner radius)^2] dy.
[Notice that the rectangle's width (dy) is now the height of our cylinder]

Note your outer and inner radii need to be in terms of y. So, they are the x-values of your functions at a given y.
y=x^2 ==> x = sqrt(y)
y = sqrt(x) ==> x = y^2

It's clear from the picture that inner radius is x=y^2 and outer is x=sqrt(y), so you get

Integral from 0 to 1 of
[pi*(sqrt(y))^2 - pi*(y^2)^2] dy,
which gives 1/2 - 1/5.

The second example is doable this way as well. Draw a picture and find outer and inner radius -- in this case when we try to solve for x in terms of y, we'll get the solutions to a quadratic.

I'd recommend, though, doing the second example using the "shell method", as others have suggested. In contrast to the "washer method" where our rectangles ran perpendicular to our axis of revolution, in the shell method, our rectangles are parallel to our axis of revolution. When we revolve them (think about revolving one of them), we get what looks like a piece of paper rolled so its ends touch (actually the nice picture here is a roll of toilet paper -- one of these revolved rectangles will look like the cardboard tube in the middle). The volume of that tube, then, is 2*pi*r*h*dx (where dx is the very little bit of thickness that the tube has, r is the distance to the axis of revolution i.e. the tube's radius, and h is the height of the tube). See the nice drawing already posted for those calculations.

Hope this helps.