How do you determine the inner and outer radius using the washer method? Conside

Question

How do you determine the inner and outer radius using the washer method? Consider the region R bounded by y=x^2-6 and y=x. Rotate about line y=6. No need to integrate. Just tell me the inner and outer radius thanks! 1. Do you look at the graph and see which line is further from the axis of revolution (and make that the outer radius) ? 2.Find the intersection of both functions then plug in points to see which number is bigger. As you can see, in this problem, the answers to determining the outer radius using 1 and 2 contradict each other.

Explanation / Answer

The outer radius is what part of the region is farther away from the line of rotation.

Note in the washer method, because we are squaring both the inner and outer radius

it doesn't matter if we take the two radii as the difference from the axis of revolution to the boundary curves or the difference from the boundary curves to the axis(line) of revolution.

For your question, note the two curves intersect at x = -2 and x = 3.

in between two points y = x is greater than y = x^2 - 6.

The outer radius is : R(x) = 6 - (x^2 - 6) = 12 - x^2

The inner radius is: r(x) = 6 - x

For calculation note that picking R(x) = (x^2 - 6)- 6 = x^2 - 12

r(x) = x - 6

Would still give you the right answer since:

When you square them, they give you the same thing.

I guess since we like to thing the radius as being positive

then the first answer I gave is the most logical since the radius will

be positive in the given interval.

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