All right. So, let's take a look at a second example. This says that the base of a solid is bounded by y = x cub, y = 0, and x = to 1.
Find the volume of the solid whose cross-sections taken perpendicular to the yaxis are equilateral triangles. Now to sketch it to be honest sketch the solid which is difficult. Okay then I'll just skip for now.
Now for the um region or um you know the base. Okay. Okay.
So this is y = to um x cube uh x = 1. So x = to 1 and y = 0. This is y= 0.
which means uh this is the base. Okay, so that's the base. Okay, now cross-section.
Next would be the cross-section. So the cross-section is equal to triangles. Okay.
So equilateral triangles. So all of them are side. Whoopsie.
And all the sides are the same. Okay. Now I want to emphasize that this is perpendicular.
Okay, the cross-section is perpendicular to y axis, right? To the y-axis. Meaning I have something like this here.
I have this. I have this. Those are the side.
Okay, those are the side of the triangle. And imagine that the triangle is standing. Okay.
Is standing on on the um you know on the screen. Okay. So you're looking it from the top.
Okay. [snorts] And you can see that we sum up in this direction. So this direction which means we're going to have dy.
So we sum up this direction. Right? we add them up.
Okay. And it is not hard to see that this is one one and this is 0 0. So this can also give us the bounds from zero to one.
I'm pretty much just done right now. Well, we're going to do the side later. So first thing first, we're going to set um we can see the intersections or sometimes we encode corners of the regions or 0 0 um one zeros and one one right to see it.
Then the volume of the solid V will be from 0 to one area of the cross-section always dy. Okay. And now back to this cross-section.
Okay, the cross-section are equilateral triangles. So this area would be 12 base * height or we can use the law of sign somewhere something like that. So side * side * sin 60° because this is 60°.
Okay, which is 12 side squared sin 60 would be right 3 over two and finally I have red 3 over 4 * side squared okay so that's the area of the equilateral triangle which is 01 right 3 over 4 time side square dy where side equals. So this time we need to come back here. So side would be this guy minus this guy.
So we have x right x left. Okay it's horizontal, right? Therefore the side will be x on the right minus x on the left.
So x on the right come back. [snorts] So x on the right is right here which is x = to 1. So I'm just use this x = to 1 minus and x on the left.
The x on the left would be this. Right? at this curve which is right here.
So y = x cub y = x cub meaning x = 1 - x = y/3. We solve for for x which is 1 - y to the 1/3. That's the side.
Then we're ready to solve it. 0 1 3 4 1 - 1/3 squared dy and we multiply things out. And maybe it would be a good idea to take this one out.
So 1 - 2 y^ 1/3 + y to the 2/3. Okay? Cuz y to the one cuz y to the 1/3 squared is y to the 1/3 * 2 which is y to the 2/3.
Okay. And erase it. Okay.
Dy and then figure out the anti-derivative. And we're pretty much done. Right.
3 over 4 times y - 2 y 4 over 3 which is 3 over 4 on top in the front + y to the 5 over 3 + one right and then I have 3 over 5. That's the anti-derivative and easy one because it's from 0 to 1. Therefore, the final answer would be the coefficient 1 - 3 / 2 + 3 over 5.
A little tough. Not bad. 10 again.
So, 10 - 15 + 6 or 1 over 10. Really? Uh yeah, 1 over 10.
So, that's 40. Okay, that's what we have for this problem. Okay.
So next I'm going to give you one more example. In this example we do not have any equation. Okay.
Like you have to build equation all by ourel.
