Can someone check my work on this volume bounded by curves problem?

[tastybrownies]

Homework Statement

Hello everyone, I was wondering if someone could check my work on this problem as I'm not sure it's right. I'm in Calc II right now and we are doing finding volume bounded by curves rotated around an axis.

So, here is the problem, y=cos(pix)+1, y=4x^(2)-9, x=0; about x=0

I wasn't sure if the bounds were to be 0 to the intersection point of 1.607006, or from -1.607006 to 1.607006. But I did the integration work shown below.

I used the shell method.

2pi * the integral of (x)(cos(pix)+1) - (4x^(2)-1) (this is for the height)

2pi* the integral of (xcos(pix)-4x^(3)+10x)

I used integration by parts for the xcos(pix) and got u=x, du=dx, dv=cos(pix), v=1/pi*sin(pix)

I went through doing the parts and I came up with x(1/pi*sin(pix)) - integral of VDU

I did the integral, did the integrals of 4x^(3) and 10x, put everything together and came up with

2pi* [(x)(1/pi*sin(pix)) + 1/(pi)^(2) * cos(pix) -x^(4) +5x^(2)]+C

Is this correct? Thank you very very much for whoever looks over this, I know this is a lot of work.

 

[Quinzio]

the bounds were to be 0 to the intersection point of 1.607006, or from -1.607006 to 1.607006.

What is this ?
I think it's not correct.

 

[tastybrownies]

If you graph this on a graphing calculator that is the right intersection given the lines of y=cos(pix)+1 and y=4x^(2)-9.

 

[dynamicsolo]

That certainly is a weird choice of functions, since such problems usually don't have integration limits that are transcendental numbers; for the moment, call whatever that value is $$x = \alpha .$$

Your shell integral looks correct, so we have

$$2\pi \int_{0}^{\alpha} [ x \cos(\pi x ) - 4x^{3} + 10x ] dx$$

So you'll need to evaluate

$$2\pi [ \frac{1}{\pi} x \sin(\pi x ) + \frac{1}{\pi^{2}} \cos(\pi x ) - x^{4} + 5x^{2} ] = 2 x \sin(\pi x ) + \frac{2}{\pi} \cos(\pi x ) - 2\pi x^{4} + 10\pi x^{2}$$

at $$x = \alpha ,$$ and you will only subtract $$\frac{2}{\pi} ,$$ since all the other terms are zero at x = 0 .

It is probably reasonable to ask at this juncture if you're sure you copied the functions in the problem correctly. If so, I think you're going to have to just enter "alpha" as the value, since there isn't going to be an expression for it that you can write down. Alternatively, if approximate values for the volume are acceptable, you could "plug in" your result of 1.607 to get a numerical result for the volume.

 

[tastybrownies]

Yeah, I did the integration and came up with the same thing you did, plugged in the 1.6 number and 0 for FTC part 2, and came up with 24. something. Someone else in the class got the same thing that I did, so it's starting to look correct. And on the crazy, bounds of integration, this is somewhat of a review for our final exam, but I think the teacher should actually pick easy to work with numbers. I mean, what experience is gained by working with number like 1.600005465465447787 instead of 2?

 

[dynamicsolo]

Yes, your exam will probably use numbers that are easier to work with, particularly if you are not allowed to use a calculator. A "real-world" problem could well involve values which can only be approximated, but then in physics or engineering, we're used to dealing with situations like that...