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We've seen that we can use either horizontal strips or vertical strips when

using integrals to calculate area. We end up computing the same number, the

area of this region, whether we use vertical rectangles or horizontal

rectangles, we're getting the same answer, the area of this region.

But if both work, which one should we choose?

The best choice depends on the shape of the region.

I like to think about how many different kinds of edges my rectangles will touch.

Here is an example to demonstrate what I mean.

I've drawn some region in here with three edges, this curved edge, this curved

edge, and this straight edge along the x axis.

Now, I could calculate the area of the enclosed region.

Either by decomposing this into vertical strips or horizontal strips.

The bad news about vertical strips is that there's two different kinds of

vertical strips. There's vertical strips over here, and

there's vertical strips over here. The vertical strips over here touch the

orange and the purple edge, and the vertical strips over here would touch the

orange and the blue edge. Contrast that with the much better

situation that happens if I cut this up into horizontal strips.

All of my horizontal strips have an orange side and a blue side.

I only have one type of horizontal strip. So I should calculate the area of a

region like this using horizontal strips. Let's take this advice and apply it to

solve a specific problem. So here's a bigger, specific version of

this problem. Let's figure out the area of this region.

this orange curve is the curve y equals the square root of x, and this blue curve

is y equals the square root of 2x minus one.

And then here I've just got the purple line along the x-axis.

I've really got a choice as to how to approach this.

I could do this with vertical strips. But then I'd have to do two different

integrals. I'd have to handle the case over here

where rectangles are touching the orange line and the purple line.

And I'd have to do an integral over here where my rectangles are touching the

orange line and the blue line. It's going to be better to use horizontal

strips. In that case, all of the rectangles are

the same kind of rectangle, they're a rectangle with one edge on this orange

curve, and one edge on this blue curve. Let me just draw one of those thin

horizontal rectangles. I have to figure out the size of that

rectangle. Let's suppose this thin strip is at

height y. I know how tall this strip is.

I'll call that dy. But how wide is this green strip?

I should figure out what this point and where this point is.

And if I think about it a little bit, if this is y, I can solve this equation and

find out that in that case x. Is y squared, and I can similarly solve

this equation and find out that if this point has a y coordinate y, the x

coordinate will be y squared plus 1 over 2.

And that tells me the width of this green rectangle, it's this minus this.

I'm going to write down the area of that green rectangle.

So the width is y squared plus 1 over 2 minus y squared, that's how wide this

rectangle is. And the height of that rectangle is dy,

so this product is the area of that green rectangle.

But of course I don't care just about that one rectangle, I want to integrate

so that I'm adding up the areas of all kinds of thin rectangles.

And then in the limit I'm getting the area of this region exactly.

So instead of just this single rectangle, I'm going to integrate y goes from where

to where. Well, I look at my picture here, y can be

as small as zero and as tall as this point up here which is one.

So y goes from zero to one and this interval calculates the area of that

region. Let's do the calculation.

This integral is the integral from 0 to 1.

for I can simplify the integrand a bit. It's 1 half minus y squared over 2 dy .

And now we can write an antiderivative for that.

So I add a derivative of 1 half. One half y and a derivative of y squared

over 2 is y cubed over 6. I'm going to evaluate that at 0 and 1,

take a difference. But when I plug in 0 I just get 0.

So the answer is whatever I get when I plug in 1.

And that's 1 half times 1, minus 1 cubed over 6.

Could simplify that a bit, that's 1 half minus a 6th, and 1 half, well, that's 3

6ths, so 3 6ths minus 1 6th, is 2 6ths, which you might write as, 1 3rd.

So the area of that, shark fin shaped region is a third of a square unit.

Often, it's not there are right ways or wrong ways but there are better ways and

worse ways, right. Ways that are better in that they are

quicker, safer, more elegant. So don't just make good choices.

Make the best choices.