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Welcome back.

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In the last video we were just figuring out the volume under

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the surface, and we had set up these integral bounds.

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So let's see how to evaluate it now.

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And look at this.

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I actually realized that I can scroll things, which is quite

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useful because now I have a lot more board space.

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So how do we evaluate this integral?

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Well, the first integral I'm integrating with respect to x.

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Right?

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I'm adding up the little x sums.

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So I'm forming this rectangle right here.

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Right?

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Or you could kind of view it I'm holding y constant and

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integrating along the x-axis.

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I should switch colors.

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So what's the antiderivative of x y squared with respect to x?

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Well it's just x squared over 2.

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And then I have the y squared-- that's just a

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constant-- all over 2.

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And I'm going to evaluate that from x is equal to 1 to x is

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equal to the square root of y, which you might be daunted by.

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But you'll see that it's actually not that bad

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once you evaluate them.

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And then let me draw the outside of the integral.

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This is y is equal to 0 to y is equal 1.

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dy.

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Now, if x is equal to 1 this expression becomes

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y squared over 2.

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Right?

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y squared over 2, minus-- now if x is equal to square root

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of y, what does this expression become?

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If x is equal to the square root of y, then

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x squared is just y.

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And then y times y squared is y to the third.

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Right?

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So it's y to the third over 3.

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Fair enough.

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And now I take the integral with respect to y.

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So now I sum up all of these rectangles in the y direction.

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0, 1.

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This is with respect to y.

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And that's cool, right?

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Because when you take the first integral with respect to x you

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end up with a function of y anyway, so you might as well

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have your bounds as functions of y's.

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It really doesn't make it any more difficult.

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But anyway, back to the problem.

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What is the antiderivative of y squared over 2 minus

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y to the third over 3?

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Well the antiderivative of y squared-- and you have to

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divide by 3, so it's y cubed over 6.

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Minus y to the fourth-- you have to divide by 4.

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Minus y to the fourth over-- did I mess up some place?

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No, I think this is correct. y to the fourth over 12.

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Oh wait.

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How did I get a 3 here?

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That's where I messed up.

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This is a 2, right?

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Let's see.

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x is square root of y.

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Yeah, this is a 2.

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I don't know how I ended up.

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Right?

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Square root of y squared is y, times y squared

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y to the third over 2.

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Right.

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And then when I take the integral of this

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it's 4 times 2.

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8.

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Got to make sure I don't make those careless mistakes.

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That's the tough part.

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I just want to make sure that you got that too.

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I hate it when I do that.

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But I don't want to re-record the whole video.

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So when I evaluated this-- right.

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This is right, and then I take the antiderivative of y to the

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third over 2, I get y to the fourth over 8.

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And now I evaluate this at 1 and 0.

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And that give us what?

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1/6 minus 1/8 minus-- well both of these when you evaluate

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them, are going to be 0's.

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So this is going to be another 0 minus 0.

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So you don't have to worry about that.

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So what's 1/6 minus 1/8?

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Let's see.

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24.

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That's 4 minus 3 over 24, which is equal to 1/24-- is the

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volume of our figure.

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So this time, the way we just did it, we took the integral

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with respect to x first, and then we did it with

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respect to y.

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Let's do it the other way around.

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So let me erase some things.

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And hopefully I won't make these careless mistakes again.

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I'll keep this figure, but I'll even erase this one.

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Let me erase all of this stuff.

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We have room to work with.

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So I kept that figure.

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But let me redraw just the xy plane, just so we get

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the visualization right.

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It's more important to visualize the xy plane in

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these problems than it is to visualize the

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whole thing in 3-d.

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That's the y-axis, that's the x-axis.

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OK.

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Our upper bound, you can say, is the graph y is

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equal to x squared.

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Or, you could view it as the bound x is equal to

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the square root of y.

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That is x is equal to 1, that's y is equal to 1.

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And we care about the volume above the shaded region.

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Right?

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That shaded region is this yellow region right here.

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And let's draw our da.

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I'll draw a little square actually.

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And I'll do it in magenta.

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So that's our little da.

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And the height is dy.

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That's a y.

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dy.

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And it's with this dx.

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Right?

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So the volume above this little square-- that's the same

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thing as this little square.

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Just like we said before, the volume above it is equal to

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the value of the function.

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Right?

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The height is the value of the function, which is x y squared.

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And then we multiply it times the area of the base.

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Well the area of the base, you could say it's da, but we know

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it's really dy times dx.

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I didn't have to write that times there.

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You can ignore it.

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And I wrote the y first just because we're

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going to integrate with respect to y first.

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We're going to sum in the y direction.

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So what does summing in the y direction mean?

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It means we're going to add that square to that

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square to that square--

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[COUGHING]

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Excuse me.

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So we're adding all the dy's together, right?

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So my question to you is, what is the upper bound on the y?

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Well once again, we bump into the curve.

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Right?

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So the curve is the upper bound when we go upwards.

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And what is the upper bound on the curve?

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Well we're holding x fixed, so for any given

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x what is this point?

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Well that's going to be x squared.

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Right?

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Because this is the graph of y equals x squared.

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So our upper bound is y is equal to x squared.

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And what's our lower bound?

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We can keep adding these squares down here.

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We're adding all the little changes in y.

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So what's our lower bound?

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Well our lower bound is just 0.

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That was pretty straightforward.

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So this expression, as it is written right now, is the

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volume above this rectangle.

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Right?

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Let me draw it.

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Is the volume above this rectangle.

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These are the same rectangles.

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The volume above that rectangle.

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Now what we want to do is add up all the dx's together, and

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we'll get the volume above the entire surface.

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So that rectangle, now we'll add it to another

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dx one, dx one like that.

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So what's the upper and lower bounds on the x's?

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We're going from x is equal to 0, right?

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We don't bump into the graph when we go all the way down.

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So we go from x is equal to 0.

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And then our upper bound is x is equal to 1.

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So x is equal to 0, x is equal to 1.

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And in general, one way to think about it is when you're

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doing kind of the last sum or the last integral, you really

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shouldn't have variable boundaries at this point.

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Right?

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Because our final answer has to be a number, assuming

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that we're not dealing with something very, very abstract.

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But our final answer is going to have a number.

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So if you had some variables here you probably

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did something wrong.

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And it's always useful, I think, to draw that little da

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and then figure out-- OK, I'm summing to dy first.

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When I go upwards I bump into the curve.

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What is that upper bound if x is constant?

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Oh, it's x squared. y is equal to x squared.

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If I go down I bump into the x-axis, or y is equal to 0.

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And so forth and so on.

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So now let's just evaluate this and confirm that

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we get the same answer.

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So we're integrating with respect to y first.

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So that's x y to the third, over three.

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At x squared and 0.

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And then we have our outer integral. x goes from 0 to 1.

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dx.

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If we substitute x squared in for y.

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x squared to the third power is x to the sixth.

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x to the sixth times x-- let me write that.

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So we have x times x squared to the third power, over 3.

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That equals x to the seventh.

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Right? x squared to the third-- you multiply the exponents, and

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then you add these. x to the seventh over 3, minus this

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evaluated with y is 0.

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But that's just going to be 0.

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Right?

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And then we evaluate that from 0 to 1, dx.

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We're almost there.

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Increment the exponent.

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You get x to the eighth over 8.

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And we already have a 3 down there, so it's over 24.

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And you evaluate that from 0 to 1.

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And I think we get the same answer.

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When you evaluate it at 1 you get 1/24 minus 0.

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So once again, when we integrate in the other order

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you still get the volume of the figure, being 1/4,

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whatever, cubic units.

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Anyway, see you in the next video.

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