Find the volumes of the solids. The base of a solid is the region bounded by the graphs of y=3 x

Find the volumes of the solids. The base of a solid is the...

Key Concepts

Volume by Cross Sections

This concept involves finding the volume of a solid by integrating the areas of cross sections taken perpendicular to a given axis. It is based on the idea that the solid can be thought of as being made up of many thin slices, each with a small volume, and the total volume is obtained by summing these slices using integration.

Definite Integration for Volume Calculation

Definite integration is used to sum up the contributions of the individual cross-sectional areas over a specific interval. By setting up an appropriate integral with the correct limits, one can compute the total volume of the solid, which is a fundamental application of integral calculus in geometry.

Cross-Sectional Area Functions

Determining the area of each cross section as a function of its position is a key step in solving volume problems by slicing. This involves expressing the dimensions of the slice in terms of the variable of integration and applying the appropriate geometric area formulas, which is crucial for setting up the integral accurately.

Determining Integration Limits

Identifying the correct limits of integration is essential for accurately computing volumes. These limits often come from the points of intersection of the curves defining the boundary of the base region, ensuring that the integration covers the entire solid without omission or redundancy.

Transcript

00:01 We want to find the volume, given the base of this solid, is bounded by the graphs of y is equal to 3x, y is equal to 6, and x is equal to 0.

00:16 And it has cross sections that are perpendicular to the x -axis that are rectangles.

00:23 And in part a, they will all have a height of 10.

00:28 And in part b, the rectangles will have a perimeter of 20.

00:36 So let's go ahead and do 10 at first.

00:43 So let's just kind of draw a little graph of what these will look like.

00:47 So let's say we have this point here, and we're going to go to here, and then we're just going to go straight up with some kind of rectangle.

01:01 And at least in part a is going to be really...

01:06 Simple for us because we already know what the height is and all we really need to figure out is what our bounds of integration are going to be as well as what is this length right here so i'll call that s so first let's just do that off on the side to find s well s is going to be well it touches this blue line where it's y is equal to six so six minus whatever the point is on this red line and the red line is 3x.

01:48 So to find that distance right there, we would have just s is equal to 6 minus 3x.

01:59 And then to find what our bounds of integration are going to be, well, we know we're starting at zero because remember also it tells us that let me do this in blue.

02:17 We're bounded by x is equal to zero, which is this line here.

02:20 X is equal to zero.

02:24 And what we need to find is what this point of intersection is.

02:28 So, well, it intersects to line y is equal to six.

02:33 So to find that point, we can just do six is equal to 3x, divide each side by three, and get two is equal to x.

02:43 So this tells us that our bounds of integration, we're going to.

02:47 Going to be 0 to 2.

02:51 So just off on the side.

02:53 So we have 0 2.

02:56 And then we have some area function, a of x, dx.

03:04 So at least in part a, like i was saying, is going to be pretty straightforward.

03:09 Since we know that the height of all these are 10, and we know this side length here, so maybe i should have named that l.

03:21 Instead, i'll just keep saying this s, will be 6 minus 3x, and we know the area of a triangle is just its height times its other side.

03:33 So it will be 10, 6 minus 3x.

03:40 So we can go ahead and take this and plug it in to our integral that we have.

03:47 And doing that will give us, so we have the integral of 0 to 2.

03:56 And remember the one of the nice things about integration is if i have a constant out here, i can really just move it out front.

04:03 So i'm just going to put that 10 out there.

04:06 And then i want to integrate 6 minus 3xx.

04:15 And i can integrate each of these using the power rule for integration.

04:23 So this is going to be 10.

04:27 So 6 integrates to 6x, and then minus 3, and then it's going to be x to the second power, since it was before to the first, and then i need to divide by the new power.

04:48 I'll put that back in blue, and it's going to be evaluated from 0 to 2.

05:01 And just to kind of simplify this a little bit more, i'm going to factor out a 3x.

05:07 So i can write this as 30x2 minus 1 half x.

05:21 All right...

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