Washer method for finding volume. Is my answer correct?

  • Thread starter Lo.Lee.Ta.
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  • #1
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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! :)
 

Answers and Replies

  • #2
Simon Bridge
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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.
 
  • #3
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...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...
 
  • #4
Simon Bridge
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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.
 
  • #5
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Am I right so far to say that R = 2 and r = 2 - √(1 - x2)...?
 
  • #6
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So, people. Would you please let me know if my R and r values are right?

...Please?

Thank you.
 
  • #7
Simon Bridge
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Yes - the limits look fine now that I see what the problem is saying.
How did you evaluate the integral?
 

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