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### Course: Calculus, all content (2017 edition) > Unit 6

Lesson 9: Disc method- Disc method around x-axis
- Generalizing disc method around x-axis
- Disc method around y-axis
- Disc method rotation around horizontal line
- Disc method rotating around vertical line
- Calculating integral disc around vertical line
- Disc method: revolving around x- or y-axis
- Disc method worksheet

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# Disc method worksheet

Gain some risk-free experience with applying the disc method in this worksheet, before you attempt our exercise.

## Problem 1

## Problem 2

## Problem 3

## Problem 4

## Problem 5

## Problem 6

## Want to join the conversation?

- For problem 6, isn't the radius of the disc (9 - y), not (9 - y^2)? The explanation says you can "see this from the graph" but what I see is [9 - f(x)] and f(x) = Y; therefore, r = (9 - y), right?(3 votes)
- f(x)=√x, not y. You should be integrating π∫[9-f(y)]dy. The given curve is y=√x, but if we're going to evaluate an integral using dy then our equation also has to be in terms of y. If we square both sides of y=√x, we get y²=x. So f(y)=y², and our radius r=9-f(y) is 9-y²

hope this helps ♥(8 votes)

- consider this problem "y=4-x^2 bounded by y=0 rotated around y=4 What is the volume?"(2 votes)
- why are the bounds y=0 and y=3.... how did you find y=3(1 vote)
- For # 6, why is the radius of the disk (9-x) instead of (9-sqrtx) ?(0 votes)