# Washer method for finding volume. Is my answer correct?

• Lo.Lee.Ta.
In summary, the conversation is about finding the volume of a half-torus obtained by rotating the unit disk above the x-axis around y=2 using the washer method. There is some uncertainty about the answer and the correct values of R and r. However, after clarifying the situation, it is determined that the limits and radii are correct and the conversation ends with a request for help in evaluating the integral.
Lo.Lee.Ta.
1. Consider the torus obtained by rotating the unit disk enclosed by x^2 + y^2 = 1 about the horizontal axis
y = 2.
Using washers, carefully find the volume of the half-torus obtained by revolving the part of the unit
disk above the x-axis around y = 2.

2. First of all, am I even setting this up properly?
I am using the washer method to solve. This is how I thought to do it...
But when I put the integral problem I get into Wolfram Alpha, it comes up with a different answer. Is my integration really wrong?

R= 2
r= 2 - √(1 - x^2)

∫-1 to 1 of [∏(2)^2 - ∏(2 - √(1 - x^2))^2]dx

=∫ ∏[-4(1 - x^2)^(1/2) + 1 - x^2]

= ∏[-4 * 2/3(1 - x^2)^3/2 * -1/2x + x - ((x^3)/3) |-1 to 1

= Substituting... ∏(1 - 1/3) - ∏(-1 + 1/3)

= 4∏/3 <-----That's my answer... But when I put my integral into Wolfram Alpha, it says the answer is 15.5504...

Ugh. Would you please tell me what I'm doing wrong?
Thank you! :)

You seem to be uncertain about your answer - which part of the working has you the most worried? (i.e. what is the reason you are so uncertain?)

The way to check is to go back and write out notes to show how you are thinking at each stage. It helps to sketch the situation too.

I think you have made the outside radius of the washers a constant with x.
Is this constant?

-----------------
Note: "∏" is upper-case pi and usually stands for a product - the lower case pi is "π" which is between the λ and the ρ on the table. However, it is usually easier to just write ##\pi## instead.

...But wouldn't the outside radius be constant?

I think the directions tell me to split x^2 + y^2 = 1 into 2 different volume problems.

They first want to know the volume of the semicircle above the x-axis when it's rotated around y=2.

They next want to know what the volume is of the semicircle below the x-axis when it's rotated around y=2.

So R (outer radius) seems like it should be 2.
r (inner radius ~ which I am pretty sure represents the space in the middle) seems like it should still be 2 - sqrt(1 - x^2)...

Because to find the inner radius, you have to take the outer radius and subtract the parabola in order to be left with only the space in the middle.

*Ha! Didn't know "∏" wasn't a square root! I don't know how to write a real square root on this forum...

So it says that the torus is centered on (x,y,z)=(0,2,0) ... "the part of the unit disk above the x axis" is, therefore, on the inside of the torus. That makes sense.

To write a square root, use "√" ... but it is better to use LaTeX since ##\sqrt{1-x^2}## is better than √(1-x2) ... to see how to write the latex, click on the "quote" buttun attached to the bottom of this post and you'll see how I did it. The LaTeX parts are enclosed in double-# or double-\$ marks.

So you would be doing: $$dV = \pi (R^2-r^2(x))dx$$

Now we are on the same page :)
How did you evaluate the integral.

Am I right so far to say that R = 2 and r = 2 - √(1 - x2)...?

So, people. Would you please let me know if my R and r values are right?

Thank you.

Yes - the limits look fine now that I see what the problem is saying.
How did you evaluate the integral?

## What is the washer method for finding volume?

The washer method is a technique used in calculus to find the volume of a solid of revolution. It involves finding the integral of the cross-sectional area of the solid along the axis of rotation.

## How do I use the washer method to find volume?

To use the washer method, you need to first identify the axis of rotation and draw a representative cross-section of the solid. Then, set up an integral to find the volume by integrating the cross-sectional area along the axis of rotation.

## When is the washer method used?

The washer method is typically used when the solid of revolution has a hole or empty space inside of it. This method allows for the calculation of the volume of the solid, taking into account the empty space.

## What are the limitations of the washer method?

The washer method can only be used for solids of revolution, meaning the solid is formed by rotating a 2D shape around an axis. It also requires the cross-sectional area to be known and easily integrated.

## Is my answer correct?

Without seeing your specific answer, it is difficult to determine its correctness. However, make sure to double-check your calculations and ensure that you have properly set up the integral using the washer method.

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