# How can I write the radius equation for the disk method if the axis of revolution intersects the area between the curves

A specific example would be revolving the area between $x^2-5$ and $5x$ below the $x$-axis about $y=-2$

PS - in general, I am assuming that revolving about any other horizontal or vertical line entails subtracting the relevant variable from the constant value - is that accurate?

• It's not clear what area you are rotating.... – David Quinn Sep 13 '15 at 20:13
• There is no region between $y= x^2$ and $y=5x$ that lies below the $x$-axis? You're not rotating any region? – user860374 Sep 13 '15 at 20:15
• He's rotating the region between $x^2$ and $5x$ from $x\in [0,5]$ about $y=-2$. – zahbaz Sep 13 '15 at 20:16
• @zahbaz - He stated "the area between $x^2$ and $5x$ below the $x$-axis "? There is no such region. I agree that I think he meant what you said, but his wording was incorrect. – user860374 Sep 13 '15 at 20:17
• sorry - typo fixed – Joe Stavitsky Sep 13 '15 at 20:22

Start with drawing a picture. In this case, think washer or donut instead of disk. But first, think of how you'd calculate the area of a donut. You'd say

$$A = \pi R^2 - \pi r^2$$

where $R$ is the outer and $r$ is the inner radius. This problem is similar, except we're stacking all these washers as we march from $x=0$ to $x=5$ where the two curves intersect.

The outer radius goes from $y=-2$ to the top curve, $5x$ and the inner radius goes to $x^2$. Draw this! You ought to see

$$R = 2 + 5x$$ $$r = 2 + x^2$$

$$\pi \int_{x=0}^5 R^2 - r^2 dx$$
It looks like the problem changed to the curves $x^2 -5$ and $5x$. This is now unclear. Do you want to rotate the area bounded by those two curves and the x axis, $y=0$? If so, the rotated region will overlap itself when rotated about $y=-2$ ...
• Volumes of rotation are a physical idea, and physically speaking, if we the rotate the entire bounded region completely about $y=-2$, we would rotate the region back onto itself after rotating through $\pi$. That's not a well-defined volume. So it depends on how you define the problem. Do we stop rotating after $\pi$? If not, do we select the exterior surface to define our volume? Or, do we create two volumes - one for the rotated region below $y=-2$ and one for that above? It's up to interpretation. desmos.com/calculator/uoshfogy7q – zahbaz Sep 14 '15 at 0:00