Math: Calculus...Shit

[Aperson444]

Let R be the region in the first quadrant bounded by the graphs of y = sqrt(x) and y = x/3

Here's where I actually need help: Find the volume of the solid generated when R is rotated about the vertical line x = -1.

Now what I did was I took an outer integral over the interval [-1/3, 3]. The line x = -1 intersects y = x/3 at the point (-1, -1/3). Obviously I square the integrand and multiply by pi. Is that correct? The integrands were x = y^2 and x = 3y respectively. This way I can integrate with respect to dy rather than dx.

Then when I want to find the actual volume minus the part I don't care about (only the region R is rotated), I subtract the integral (again, with respect to dy) of the same exact expression pi*integral[(y^2-3y)^2dy] except I integrate over the interval [-1/3, 0]. The region R only spans [0, 3]. Thus, I take away that tiny useless chunk and am left with the volume of the expression rotated around the axis.... Correct?

[Mantikore]

Remember, the region exists in the first quadrant. The first thing i would do is draw the two functions out. Should look like so:

http://www.wolframalpha.com/input/?i...3+from+0+to+10

as you can see, the region should be between y = 0 and y = 3

Youve changed your two equations to have it in terms of y, which is good.

Now, the volume of a revolution is the definite integral of the square of the function, between our two x values. So all you would need to do is find the volume of the 'bigger one' in this case 3y and subtract it by the volume of the smaller one.

The problem here is that the axis is at x = -1. What I would do in this case, is just mod the equations to be 3y +1 and y^2 +1, and pretend its rotating on the y axis. it should still get the right answer

if someone has something better, please chime in

[Hooky the Cripple]

I would just translate the region to the right by one unit then do a triple integral in cylindrical coordinates. I got 207pi, which seems a little high. May have fucked up somewhere...

Edit: yeah, it should be (207*pi)/5 perhaps.

[Myxomatosis]

[Aperson444]

So I would just shift the curves 1 unit first (after they are converted to a form expressing x in terms of y), then square and integrate with respect to dy over the appropriate interval of the region, which is [0, 3] in this case? I don't need to compensate for that transformation by 1 unit, right?

[Mantikore]

shouldnt be a problem if you visualise what the solid is going to look like. Do you have the correct answer for your question written somewhere that you can check?

[Fenrisulfr]

I'd be happy to help you. I need to brush up on this stuff myself.

There are two integrals to be done: one for the region bound by y = sqrt (x) + 1 and the other for y = x/3 + 1. The +1 offset is necessary since the axis is at x=-1.

The general equation is the integral from x=a to x=b is V=integral from a to b of pi*(f(x))^2. Apply this to the two functions. Take the difference of the outer volume and the inner volume and that will be the final result.