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When should I use horizontal as opposed to vertical pieces? - Week 15 - Lecture 2 - Mooculus - YouTube
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[MUSIC].
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We've seen that we can use either
horizontal strips or vertical strips when
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using integrals to calculate area.
We end up computing the same number, the
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area of this region, whether we use
vertical rectangles or horizontal
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rectangles, we're getting the same
answer, the area of this region.
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But if both work, which one should we
choose?
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The best choice depends on the shape of
the region.
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I like to think about how many different
kinds of edges my rectangles will touch.
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Here is an example to demonstrate what I
mean.
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I've drawn some region in here with three
edges, this curved edge, this curved
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edge, and this straight edge along the x
axis.
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Now, I could calculate the area of the
enclosed region.
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Either by decomposing this into vertical
strips or horizontal strips.
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The bad news about vertical strips is
that there's two different kinds of
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vertical strips.
There's vertical strips over here, and
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there's vertical strips over here.
The vertical strips over here touch the
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orange and the purple edge, and the
vertical strips over here would touch the
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orange and the blue edge.
Contrast that with the much better
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situation that happens if I cut this up
into horizontal strips.
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All of my horizontal strips have an
orange side and a blue side.
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I only have one type of horizontal strip.
So I should calculate the area of a
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region like this using horizontal strips.
Let's take this advice and apply it to
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solve a specific problem.
So here's a bigger, specific version of
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this problem.
Let's figure out the area of this region.
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this orange curve is the curve y equals
the square root of x, and this blue curve
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is y equals the square root of 2x minus
one.
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And then here I've just got the purple
line along the x-axis.
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I've really got a choice as to how to
approach this.
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I could do this with vertical strips.
But then I'd have to do two different
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integrals.
I'd have to handle the case over here
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where rectangles are touching the orange
line and the purple line.
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And I'd have to do an integral over here
where my rectangles are touching the
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orange line and the blue line.
It's going to be better to use horizontal
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strips.
In that case, all of the rectangles are
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the same kind of rectangle, they're a
rectangle with one edge on this orange
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curve, and one edge on this blue curve.
Let me just draw one of those thin
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horizontal rectangles.
I have to figure out the size of that
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rectangle.
Let's suppose this thin strip is at
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height y.
I know how tall this strip is.
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I'll call that dy.
But how wide is this green strip?
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I should figure out what this point and
where this point is.
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And if I think about it a little bit, if
this is y, I can solve this equation and
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find out that in that case x.
Is y squared, and I can similarly solve
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this equation and find out that if this
point has a y coordinate y, the x
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coordinate will be y squared plus 1 over
2.
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And that tells me the width of this green
rectangle, it's this minus this.
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I'm going to write down the area of that
green rectangle.
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So the width is y squared plus 1 over 2
minus y squared, that's how wide this
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rectangle is.
And the height of that rectangle is dy,
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so this product is the area of that green
rectangle.
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But of course I don't care just about
that one rectangle, I want to integrate
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so that I'm adding up the areas of all
kinds of thin rectangles.
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And then in the limit I'm getting the
area of this region exactly.
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So instead of just this single rectangle,
I'm going to integrate y goes from where
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to where.
Well, I look at my picture here, y can be
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as small as zero and as tall as this
point up here which is one.
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So y goes from zero to one and this
interval calculates the area of that
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region.
Let's do the calculation.
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This integral is the integral from 0 to
1.
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for I can simplify the integrand a bit.
It's 1 half minus y squared over 2 dy .
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And now we can write an antiderivative
for that.
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So I add a derivative of 1 half.
One half y and a derivative of y squared
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over 2 is y cubed over 6.
I'm going to evaluate that at 0 and 1,
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take a difference.
But when I plug in 0 I just get 0.
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So the answer is whatever I get when I
plug in 1.
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And that's 1 half times 1, minus 1 cubed
over 6.
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Could simplify that a bit, that's 1 half
minus a 6th, and 1 half, well, that's 3
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6ths, so 3 6ths minus 1 6th, is 2 6ths,
which you might write as, 1 3rd.
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So the area of that, shark fin shaped
region is a third of a square unit.
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Often, it's not there are right ways or
wrong ways but there are better ways and
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worse ways, right.
Ways that are better in that they are
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quicker, safer, more elegant.
So don't just make good choices.
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Make the best choices.
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