Volume with cross sections: squares and rectangles (no graph) | AP Calculus AB | Khan Academy

The base of a solid is the region enclosed by the graphs of ( y = -x^2 + 6x - 1 ) and ( y = 4 ). Cross sections of the solid perpendicular to the x-axis are rectangles whose height is ( x ). Express the volume of the solid with a definite integral. So pause this video and see if you can have a go at that.

All right, now what's interesting about this is they've just given us the equations for the graphs, but we haven't visualized them yet, and we need to visualize them. Or at least, I like to visualize them so I can think about this region that they're talking about. So maybe a first thing to do is think about, well, where do these two lines intersect?

So when do we have the same ( y ) value? Or another way to think about it is when does this thing equal 4? So if we set them equal to each other, we have ( -x^2 + 6x - 1 = 4 ). This will give us the ( x ) values where these two lines intersect. If we want to solve for ( x ), we can subtract 4 from both sides, and we get ( -x^2 + 6x - 5 = 0 ).

We can multiply both sides by negative 1. We will get ( x^2 - 6x + 5 = 0 ), and then this is pretty straightforward to factor. ( 1 \times 5 = 5 ) or actually, I say negative 1 times negative 5 is 5, and negative 1 plus negative 5 is negative 6. So it's going to be ( (x - 1)(x - 5) = 0 ).

These intersect when ( x = 1 ) or ( x = 5 ). Since we have a negative out front of the second degree term, we know it's going to be a downward opening parabola, and we know that we intersect ( y = 4 ) when ( x = 1 ) and ( x = 5 ). The vertex must be right in between them, so the vertex is going to be at ( x = 3 ).

Let’s actually visualize this a little bit. It’s going to look something like this. Draw it with some perspective because we have to think about a three-dimensional shape. So that's our ( y ) axis; this is our ( x ) axis. Let me draw some ( y ) values: 1, 2, 3, 4, 5, 6, 7, 8. This is probably sufficient.

Now we have ( y = 4 ), which is going to look something like this, so that is ( y = 4 ). Then we have ( y = -x^2 + 6x - 1 ), which we know intersects ( y = 4 ) at ( x = 1 ) or ( x = 5 ). So let's see: 1, 2, 3, 4, 5.

We have that point right over there: ( (1, 4) ) and then we have ( (5, 4) ). We know the vertex is when ( x = 3 ), so it might look something like this. We could substitute 3 back in here:

( y = -9 + 18 - 1 ). What is that going to be? That's going to be ( y = 8 ). So we have the point ( (3, 8) ). This is 5, 6, 7, 8—yep, right about there.

So we are dealing with a situation that looks something like this. This is the region in question, so that's going to be the base of our solid. They say cross sections of the solid perpendicular to the x-axis.

Let me draw one of those cross-sections. This is a cross-section perpendicular to the x-axis—rectangles whose height is ( x ). So this is going to have height ( x ) right over here. The height is ( x ).

Now, what is this? The width I guess we could say of this rectangle? Well, it's going to be the difference between these two functions. It's going to be the upper function minus the lower function, so that right over there is going to be ( -x^2 + 6x - 1 ) and then minus 4, which gives us the lower function.

So that could be simplified as ( -x^2 + 6x - 5 ). If we want to figure out the volume of this little section right over here, we multiply ( x ) times this and then we would multiply that times an infinitesimally small depth ( dx ).

Then we can just integrate from ( x = 1 ) to ( x = 5 ). So let's do that. The volume of just this little slice over here is going to be the base, which is ( -x^2 + 6x - 5 ), times the height ( x ), times the depth ( dx ).

What we want to do is sum up all of these, and you could imagine right over here you would have—or like right over here—you would have a cross-section that looks like this ( x ), it's now much larger; the height is ( x ).

So now it looks something like this. I'm just drawing two cross-sections just so you get the idea. So this is any one cross-section for a given ( x ), but now we want to integrate our ( x ) going from ( x = 1 ) to ( x = 5 ); ( x = 1 ) to ( x = 5 ).

And there you have it; we have expressed the volume of that solid as a definite integral. It's worth noting that this definite integral—if you distribute this ( x ), if you multiply it by all of these terms—it's very solvable. You don't need a calculator; you're just going to get a polynomial over here that you have to take the anti-derivative of in order to evaluate the definite integral.