# Volume of a Solid using Cylindrical Shells

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1. Jun 14, 2017

### lingualatina

1. The problem statement, all variables and given/known data
Find the volume of the region bounded by the curves y=3x-2, y=6-x, and the x-axis when the region is rotated around the y-axis.

2. Relevant equations
Volume using cylindrical shells: 2π∫r(x)h(x)dx

3. The attempt at a solution
I graphed the curves and then found the x-intercept of 3x-2 (which was 2/3) and the x-value of the intersection point of the two lines (which was 2). Then I set up two integrals: 2π∫x(6-x)dx from 0 to 2/3 and 2π∫x(6-x-(3x-2))dx from 2/3 to 2. When I computed and then added these, the value was 32.579, but the answer is supposed to be 33.510. Can't figure out where I went wrong: is there a problem with splitting up the integrals like that? Thanks in advance.

2. Jun 14, 2017

### haruspex

If you draw a diagram of that you will see that you have left out a triangle below the x axis and left of x=2/3.
In fact, there is no need to break it into two integrals. The x axis is not relevant. You just care about the vertical distance from the lower line to the upper line.

3. Jun 15, 2017

### Ray Vickson

The meaning of the question is not clear to me. You have taken the rotated xy region to be to the right of the y-axis, above the x-axis, above the line y=3x-2 and below the line y = 6-x, which gives you a 4-sided polygon.

When I first saw the problem it looked to me as though the xy region should be above the x-axis but below the two lines y=6-x and y = 3x-2, and that would be a triangle with vertices (2/3,0), (2,4) and (6,0).

I guess the wording also allows the region to be above the x-axis, above the line y = 6 - x and below the line y = 3x - 2, and that would be an infinite-area region extending out to x = +∞, probably not what was intended.

On the other hand, perhaps you copied the problem incorrectly. If it had said that one of the region boundaries is the y-axis (rather than the x-axis) the xy region would be as described by haruspex in post #2. That gives the required value 33.51.

Last edited: Jun 15, 2017
4. Jun 16, 2017

### Fred Wright

I assume from the problem statement that you are asked to find the solid of revolution of the triangle(0,-2),(2,4),(0,6) about the y-axis. In this problem you are stacking cylindrical discs of infinitesimal width on the y-axis. The volume of each disk is $\pi r^2dy$ where $r=x$. You have $x=\frac{1}{3}\left( y+2\right)$ and $x=6-y$. Therefore you want $$\pi\int_4^6 \left(6-y\right)^2dy+\frac{\pi}{9}\int_{-2}^4 \left( y+2\right)^2dy$$

5. Jun 16, 2017

### haruspex

Lingualatina is attempting to use cylindrical shells about the y axis, not discs. ("Cylindrical discs"??)
That is the better method here since it can all be done in one integral, though l.l. did not realise that.
Besides, this is a homework forum. Please do not post so much of the solution. Just point out errors and provide hints.

Last edited: Jun 16, 2017
6. Jun 16, 2017

### Fred Wright

My bad. I apologize. I wrote "disc" where I should have written "shell" and l went to far by writing out the explicit integrals. My technique was how I was taught to handle these problems many years ago. I thought the intent of the problem was to practice integration. It obviously has been superseded by something much better. Parenthetically, I wonder if the O.P. realizes that this problem can be quickly solved, in two lines, without the need for integration, by using the formula for the volume of a cone?

7. Jun 16, 2017

### haruspex

No, your method does indeed use stacked discs. Because the radii follow two different slopes you needed two integrals.
L.l.'s method uses cylindrical shells about the y axis, the height of each shell being the distance between the two slopes. Hence one integral does it.
Further, you can imagine sliding these shells over each other to get them aligned at one end. This reduces it to a single cone of known base area and height.