# Need help with the washer method

karamsoft

## Homework Statement

Use the washer method to find the volume of hte solid generated by revolving the regions bounded by y= root(x) and the lines y =2 and x=0 about the x-axis , y-axis , y= 2 and x=4

## Homework Equations

V= Pi integral (0,4) (R(x)^2 - r(x)^2) dx

## The Attempt at a Solution

I tried to use the equation above but I totally got confused about r(x)^2 as from what I know that the washer method means you have a small radius that you will subtract from the large curve which is in this case root(x) but there is none!

Thank you very much for your time, I would appropriate if you could explain how you solve this question so I could learn

karamsoft
I tried to solve these problems in a different way but I'm not sure if I'm on the right track.

If it revolved over the x-axis then the volume will be
integral (0,4) Pi (2)^2 - (root (x))^2 dx = 42.265 units^3

If it revolved over the y-axis then the volume will be
integral (0,2) Pi (y)^2 - 0 dy = 8.3775 units^3

If it revolved over the line y=2 then the volume will be
integral (0,4) Pi (-root(x)+2)^2 dx = 8.3775 units^3

If it revolved over the line x=4 then the volume will be
integral (0,2) Pi ((4)^2 -(-y^2+4)^2) dy = 46.9144 units^3

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I tried to solve these problems in a different way but I'm not sure if I'm on the right track.

If it revolved over the x-axis then the volume will be
integral (0,4) Pi (2)^2 - (root (x))^2 dx = 42.265 units^3

$$\int_0^4 \pi\left(2^2-(\sqrt{x})^2\right)\,dx$$

You're missing a set of parentheses.

If it revolved over the y-axis then the volume will be
integral (0,2) Pi (y)^2 - 0 dy = 8.3775 units^3
You missed the parentheses here too, but the numerical answer is OK, because zero times pi is zero.

If it revolved over the line y=2 then the volume will be
integral (0,4) Pi (-root(x)+2)^2 dx = 8.3775 units^3
This looks OK.