Calculus problem help?

[MonkeySage]

I'm supposed to use the shell method, with:
y= 4x - x2, x= 0, y= 4

Following the instructions in the text to the letter gave me:

V=2π * int(0,4)[x(4x-x2)]dx
Answer obtained from that integral was 128π/3


But, something was telling me that this was incorrect. So, I checked the back of the book.

Sure enough, I was way off.

They had:

V= 2π * int(0,2)[x(x2-4x+4)]dx =8π/3


I decided to test their solution before looking for a way to correct my own method, and the answer to their own integral is incorrect. So, if I can't even trust my own text book, I'm left with my teacher. Since it's Sunday, I have no way to get in touch with him.

[gomipile]

I'm supposed to use the shell method, with:
y= 4x - x2, x= 0, y= 4



I think that either your textbook is really bad, or you haven't provided a full statement of the problem. Perhaps you've left out some description that was at the beginning of that group of problems, or was provided in a section where the textbook explains how those problems would be set up when presented to you?

[PirateMonk]

More context would be helpful, but as far as I can tell, the problem is asking you to find the volume produced by rotating the sort of triangular region bounded by the y-axis, a particular parabola, and the horizontal line given by y=4 about the y-axis. Based on your integral, it seems like you instead found the volume produced by the region between the parabola and the line y=0. Draw a picture if you haven't already, and think about why the integral you wrote down makes sense for the region below the parabola and the integral in the book makes sense for the upper region. After that, check your work on the book's integral again; their computation looks good to me.

Hope that helps and this doesn't come too late.

[Kato]

Frankly, without the text of the exercise it's hard / impossible to judge which solution is correct...
Also, I'm not sure what the question is if you have already arrived at the conclusion that there is a mistake in your book (which I will admit is sad but since books are made by people and people make mistakes...)

[Chen]

The solution of 8*Pi/3 is correct for the integral you presented. I don't know whether or not the integral is the correct formulation since I havent used the shell method in ages, but the solution to the presented integral isn't in error.

[Knaight]

Looking at those formulas, you've got the standard 2*pi*r*h*dr in both your and the textbook's formula. Beyond that you diverge dramatically in what you're measuring.

You're using 4x-x^2 for h. They're using x^2-4x+4 for h, which can also be expressed as 4-(4x-x^2). In short, you're using the area below the line, they're using the area above the line. Then there's the bounds of the integral - you've got 0-4 for r, they've got 0-2. This makes a lot of sense for the below and above the line behavior, as the area above the line hits 0 at 2, and the area below is bounded at 0 and 4 by the x limits.

Basically, you've got the top half of a sort of toroidal structure with r=R, but where it's parabolic instead of circular. They've got a piece that fits in the middle of that structure from the top, a spike with uniformly circular cross sections.

This is why the answers are so different, and also why your answer is so much larger. Either you or the textbook solved for the wrong shape. Somehow, I suspect it's not the textbook, particularly as that y=4 line is a totally unnecessary boundary condition if you're working in the area under the curve, and that x=0 line is the axis of rotation in both cases.