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Find the volume of the described solid $ S $.

The base of $ S $ is the same base as in Exercise 56, but cross-sections perpendicular to the x-axis are squares.

$\frac{1}{3}$

Applications of Integration

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for this problem. We are looking at a solid now. The base of the solid is a triangle It has. The Vertex is 00 10 and 01 So I'm just going to shade in red. This is the base of our solid. Now. We're also told if we take cross sections perpendicular to the X axis so perpendicular to the X axis means I'm taking little cross sections just like that perpendicular. And these are squares. So you can imagine that I'm going to just be stacking squares all along the x axis. So I'm going to go from X equals 02 X equals one all along. So I'm finding the areas of squares. Well, this area of squares I'm gonna get rid of That s in just a minute, but it's going to be the side square. That's the area of each of those squares, and the height of them is going to be that infanticidal change in X. Okay, So this is that my basic setup. Now, I don't want to leave the s in there. I want to make the side in. I want to put that in terms of X. So let's take a look at our picture. My value. The height of this is going to be the value of that line. I'm gonna I'll move along. That line is I go from X equals 02 x equal one. So what is the equation of that line? Well, I can see just by examination, the Y intercept is one. Well, what's the slope? I'm going from 01 to 10 That's a slope of negative one. So that is why that's the height of that height of that curve that line as I go across. So on my integral I don't have an s. What I can do is I can say this is negative. X plus one squared. So let's square this. Yeah, that gives me X squared minus two x plus one D x. Okay, let's integrate. We can take the integral of each piece In turn. This gives me X cubed over three minus Well, that becomes X squared, divided by two. So that cancels the two that's already there. So it's just negative X squared plus x. And again, I'm integrating this from 0 to 1. So upper limit First, if I put in a one, I get one third, minus one plus one. If I put in that zero, well, every term is zero, so I can kind of ignore that piece. And when it's all said and done, the volume is one third.

Rochester Institute of Technology

Applications of Integration