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2 min read•june 11, 2020

Anusha Tekumulla

This topic is similar to topic 8.11 except we will be **revolving our shape around other axes instead of just the x- or y-axis. **These axes can be lines like x = 1, y = -2, or even y = x. This can get a bit confusing but we’ll simplify it for you with an example problem.

Suppose we have the region bounded by y = √x on top and y = x on the bottom. We are asked to find the volume of the figure if the region is revolved around the y = -1. This is the same region we used in example for topic 8.11. As you read through this consider the similarities and differences between the two procedures.

The first step of this is to **find inner and outer radii**. The outer radius will be the top function, which is √x, plus 1. We add one because we revolve around the line y = -1. The inner radius will be the bottom function, which is x, plus 1.

Now we can **calculate the area of the cross section**. This is as simple as subtracting the area of the smaller disc from the larger one. If the outer radius is (√x) + 1 and inner radius is x + 1, then the area of the washer is π((√x) + 1)2 - π(x + 1)2. Now all we have to do is find the intersections of the functions to get the endpoints for the integral. To find the intersection, set the two functions equal to each other and solve for x. In this example, the intersections would be at x = 0 and x = 1. The final step is to **integrate **to find the volume:

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