Shell method around a non-axis line 2

Calculus

Solid of revolution volume

Disc method: function rotated about x-axis
Disc method (rotating f(x) about x axis)
Volume of a sphere
Disc method with outer and inner function boundaries
Shell method to rotate around y-axis
Disk method: rotating x=f(y) around the y-axis
Shell method around a non-axis line
Shell method around a non-axis line 2

Shell method around a non-axis line 2 The last part of the problem in part 7

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Shell method around a non-axis line 2

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Where I left off, we were just essentially chugging through
this fairly hairy derivative-- this definite integral--
this antiderivative.
It takes my brain a little while to come
up with the next term.
So we evaluated at 2 and now we have to subtract
this evaluated at 1.
So minus 16 pi minus 4 pi over 3-- oh, sorry.
Plus.
The minus is going to be on everything.
Oh no, sorry.
That is a minus.
That's a minus.
Plus 4 pi.
Right?
Because x is 1.
Minus pi over 2.
And now we can simplify it.
Let's see what we can do.
This is really a hairy problem.
The 16 pi minus 8 pi, that equals 8 pi.
And then that's a plus.
The 32 plus 8 pi.
32 pi plus 8 pi, that equals 40 pi.
So let's see, I've simplified it to 40 pi, and what's minus
8 times 4 is 32 pi over 3.
And then all of that, let's see if I can simplify this.
Let's see, 16 pi plus 4 pi, that's 20 pi.
And then, let's see.
Minus 4 pi over 3 minus pi over 2.
So let's get a common denominator for this
part right here.
So if I put everything over 6-- 20 pi over 6 is the same thing
is 120 pi over 6, and then minus-- 4 pi over 3, if I put
it over 6 it becomes 8 pi over 6, right?
And then pi over 2, if I put it over 6 it becomes 3 pi over 6.
So minus 3 pi.
So this whole term that we're going to have to subtract from
this term is equal to 120 minus 11, right?
So that's 109 pi over 6.
This equals 1-- is that right?
Yeah, 109 pi over 6.
See what happens when you make up problems on the fly?
You get ugly numbers.
109 pi over 6.
And what does that top part translate?
So this is what we're going to subtract.
This is when we evaluated the antiderivative at 1.
And let's simplify this one.
So that one, if we put 3 common denominator, that's 120 pi
over 3 minus 32 pi over 3.
120 minus 32, let's see, we get 90-- 88?
Right.
88 pi over 3.
So that equals 88 pi over 3, and remember, that's
just the top part.
And then-- this is more of a review of fractions than
anything else-- and then if I want to put it over
6, I just double it.
So I think we're almost done.
Let me switch colors.
So if you go over a denominator of 6-- 88
pi over 3-- let's see.
If I double that, I get 160-- 176, right?
176 pi minus this.
Minus 109 pi.
I'm sure I made a careless mistake.
These are my least favorite type of things to do.
Hairy fractions.
So 176 minus 109.
That's the same thing as 76 minus 9.
And that is 65.
So our final answer is 65 pi over 6.
Which isn't as hairy as it seemed when we
started the problem.
But that's pretty neat.
This is kind of a strange shape.
It's kind of a wide ring that has-- the upper part of the
ring and the inner part of the ring are hard edges.
But then it's curved on the outside.
And we were able to figure out the volume of that.
Especially-- and what was weird about this, is when it was
rotated around the line y is equal to minus 2.
Hopefully I didn't confuse you.
These are about as difficult as these volume of
revolution problems get.
If you want more, just let me know.
I will see you in the next video.

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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...

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At 2:33, Sal says "double bonds" but should say "single bonds."

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