# Finding the Volume of an Oddly Shaped Region

I am currently working on some homework for my calculus course and I have been stuck on a problem for quite some time. The problem sounds simple enough:

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y=12.

• y = x
• y = 11
• x = 0

So the graphs should look like this:

Now, based on the question, I believe the object is: the area between Y = 11 and Y = 12; and the object's domain is restricted between X = 0 (y-axis) and X = 12.

Because the question states the object revolves around Y = 12, the radius should be r = 1 (12-11) except when X > 11, in which case r = 12 - X. (This accounts for the diagonal line created by Y = X.) This all seems fine to me, but I cannot seem to figure out how to write up the formula for this.

I am trying to use the disk method to solve this and I have tried many times to write up the equation. This is my most recent attempt:

$$V = \pi \, [\int_0^{11} (1)^2 \, dx + \int_{11}^{12} (12-x)^2 \, dx]$$

When that's all solved out, I end up (34*pi)/3. However, this is a multiple choice problem and the answer I am ending up with does not match any solutions.

The answer should be one of the following:

• (2057/3) * pi
• (2057/6) * pi
• (847/3) * pi
• (1694/3) * pi

Any and all help is appreciated. I would also appreciate information about solving similar problems.

Edit: To clarify, I think the question is asking for the volume of the region between (0, 11), (0, 12), (12, 12), and (11, 11).

The area bounded is a triangular region with vertices at $(0,0),(11,11),(0,11)$.

We want

$$\int _0^{11}\pi r^2dx=\int_0^{11}\pi([12-x]^2-(12-11)^2)\,dx.$$

But why? We are interested in two solids of revolution. The first one is formed by rotating $y=x$ around $y=12$. This yields the integral $\int_0^{11}\pi \times\text{radius}_1^2(x)\, dx$, where radius$_1$ is the distance from the axis of rotation to the edge of the solid, please verify that this is $12-x$. This is correct since we know that the distance when $x=0$ must be $12$ and it is, and also the distance when $x=11$ must be one, and $12-11=1$.

Now once that we calculated this volume subtract from it the volume of a smaller solid generated by rotating $y=11$ around $y=12$. Here the distance is always $1$, that's why I wrote $12-11$. So that integral should be $\int\pi\times\text{radius}_2^2(x)\, dx$. When we take the last two integrals and subtract them we obtain the first integral.

You can look it better here:

• My apologies, but can you explain how this accounts for the region between (0, 11), (0, 12), (12, 12), and (11, 11)? I believe the question is asking for the volume of that region and I do not understand how this integral finds that region. In particular, I am confused about the minimum and maximum you used on the integral. – Spencer D Jan 19 '15 at 21:36
• @SpencerDoak Unfortunately I can't see images so I can't see what you uploaded. I will upload a picture and you tell me if you can see it, ok? wait for me. Also reload the page since I edited my answer and explained it better. – Vladimir Vargas Jan 19 '15 at 21:38
• @SpencerDoak reload the page again, please tell me if you can see the image. – Vladimir Vargas Jan 19 '15 at 21:44
• Yes, thank you, I can now see the image. I am going to read through the explanation and the answer once more before I accept it as correct. Thank you for your patience and willingness to explain. – Spencer D Jan 19 '15 at 21:50
• @SpencerDoak that's alright. I'm sorry I can't see images by imgur (not even the one I posted! haha) because I could have helped you better if I saw your image. – Vladimir Vargas Jan 19 '15 at 21:52