Finding volume with the disc method, calculus help.

whatever

Question: find the volume of x = -y^2 + 4y , bounded by the y-axis and y=1, revolving around the y-axis. I keep getting (183pi)/5 but the answer is supposed to be (153pi)/5. Am I wrong or is the text book answer wrong?

GreyHuge

I think you are both wrong. I don't know what the hell either of you did, but I get V=5*Pi/3 (for y <= 1) or V=9*Pi (for y >= 1). I think you might need to spend a little more time in your post explaining what the problem is.

lackofcheese

Question: find the volume of x = -y^2 + 4y , bounded by the y-axis and y=1

The y-axis and y=1 are perpendicular to one another. It doesn't make sense to say that those are the bounds.

GreyHuge

Question: find the volume of x = -y^2 + 4y , bounded by the y-axis and y=1

The y-axis and y=1 are perpendicular to one another. It doesn't make sense to say that those are the bounds.

My thoughts exactly, and yet here we are.

lackofcheese

Am I wrong or is the text book answer wrong?

Neither. The question is wrong.

Dr. Timo

The question is not wrong, but the correct answer is (53 Pi)/15 and I suspect a typo (the one from the fifteen somehow slipped into the numerator). Switching the place of x and y in the problem (because that's the canonical way of looking at functions. you have the function

y = -x^2 + 4x
y = 0 (the x-axis)

These bound an area from the top and bottom respectively when 0 < x < 1, the lower bound is where y = -x^2 +4x crosses the x-axis and the upper bound is set by the problem.

Now this is rotated around the x-axis (remember I switches all the y <--> x in the problem). This means you'll get discs with the radius indicated by the hight of the curve in the picture. The area of these discs is

A(x) = Pi r^2 = Pi (-x^2 + 4x)^2

Now all you do is sum up all the disc from x = 0 to x = 1, i.e., integrate A(x). The result is (53 Pi)/15 (at least according to Mathematica).

Barring any embarrassing mistakes on my part, this is how it should be done.

lackofcheese

The question is not wrong, but the correct answer is (53 Pi)/15 and I suspect a typo (the one from the fifteen somehow slipped into the numerator). Switching the place of x and y in the problem (because that's the canonical way of looking at functions. you have the function

No, the question is wrong because it's ambiguous. The problem with the question is that, as Asnabel said, you could also consider the function for y > 1, which in fact gives (153 Pi)/5

Dr. Timo

No, the question is wrong because it's ambiguous. The problem with the question is that, as Asnabel said, you could also consider the function for y > 1, which in fact gives (153 Pi)/5

Indeed, I hadn't thought of that, what an interesting coincidence.

In any case my earlier analysis stands, but in the last step you just move the integration bounds to be from x = 1 to x = 4 (where the function y = -x^2 +4x crosses the x-axis again).

GreyHuge

So, using the disk method:
Evaluate from 0 to 1: 53*Pi/5
Evaluate from 1 to 4: 153*Pi/5
Ignore my earlier posting, I forgot to square the radius.
*whistles while tapping earlier post under the carpet with his foot*

whatever

There's a graph that goes with the question that makes it less ambiguous, it was late at night and I did not feel like scanning the text book. It looks like I just made an addition error and kept overlooking it or something, getting 153pi/5 now.

HungryJoe

Threads like this make me wish that the forum had an equation editor.

WindUpBird

Isn't there a LaTeX plug-in for Vanilla?

lackofcheese

That would be good. Otherwise, at least there's stuff like this.