Calculus what is a cross section

Cylinders do not change their radius in the middle of a problem and so as we move along the center of the cylinder i. In these examples the main issue is going to be determining what the cross-sectional areas are. To determine this, consider the figure on the right above. In other words, we know that,. Before we proceed with some more complicated examples we should once again remind you to not get excited about the other letters in the integrals.

The area of this disk is then,. In particular look at the triangle POR. On the left we see how the wedge is being cut out of the cylinder. The sketch in the upper right position is the actual wedge itself. Note as well that this is the reason for the way we oriented the axes here. We can also compute the volume of solids of revolution that have a hole in the center. The general principle is simple: compute the volume of the solid irrespective of the hole, then subtract the volume of the hole.

One can generate a solid of revolution with a hole in the middle by revolving a region about an axis. This leads us to the Washer Method. They help us establish the outside and inside radii. Thus the volume is. This section introduced a new application of the definite integral. Putting it all together, the following is obtained:. Explanation : First, the cross sections being perpendicular to the axis indicates the expression should be in terms of.

Explanation : Because the disk is of radius , the base is defined by the following formula:. The correct formula for the area of an equilateral triangle is as follows: , with s being the side length of the triangle. Explanation : Since the cross-sections are perpendicular to the axis, the volume expression will be in terms of. Putting this all together, we find the following:. Explanation : The base is defined by the following formula:.

The radius defines the bounds as being The correct formula for the area of a semicircle is as follows: , with r being the radius of the semicircle. This can be simplified: Putting this all together, we find the following:. Explanation : The cross-sections are parallel to the axis; this is another way of saying the cross-sections are perpendicular to the axis. Copyright Notice. View Tutors. Les Certified Tutor. Princeton University, Bachelor in Arts, Economics.

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Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure b , and using a proportion, since these are similar triangles, we have.

Then we find the volume of the pyramid by integrating from step. Use the slicing method to derive the formula for the volume of a circular cone. Use similar triangles, as in Figure.

If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution , as shown in the following figure. Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.

Use an online integral calculator to learn more. Use the slicing method to find the volume of the solid of revolution bounded by the graphs of and rotated about the. Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval as shown in the following figure. Next, revolve the region around the -axis, as shown in the following figure.

Since the solid was formed by revolving the region around the the cross-sections are circles step 1. The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by Use the formula for the area of the circle:. The volume is. Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function and the over the interval around the See the following figure.

Use the problem-solving strategy presented earlier and follow Figure to help with step 2. When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks.

To see this, consider the solid of revolution generated by revolving the region between the graph of the function and the over the interval around the The graph of the function and a representative disk are shown in Figure a and b. The region of revolution and the resulting solid are shown in Figure c and d. We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that. The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle.

This gives the following rule. Let be continuous and nonnegative. Define as the region bounded above by the graph of below by the on the left by the line and on the right by the line Then, the volume of the solid of revolution formed by revolving around the is given by. The volume of the solid we have been studying Figure is given by. Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of and the over the interval around the.

The graphs of the function and the solid of revolution are shown in the following figure. The volume is units 3. Use the procedure from Figure.

So far, our examples have all concerned regions revolved around the but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that has been generated by revolving a region around the The mechanics of the disk method are nearly the same as when the is the axis of revolution, but we express the function in terms of and we integrate with respect to as well.

This is summarized in the following rule. Define as the region bounded on the right by the graph of on the left by the below by the line and above by the line Then, the volume of the solid of revolution formed by revolving around the is given by. Let be the region bounded by the graph of and the over the interval Use the disk method to find the volume of the solid of revolution generated by rotating around the.

Figure shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the the disks are horizontal, rather than vertical. The region to be revolved and the full solid of revolution are depicted in the following figure.

To find the volume, we integrate with respect to We obtain. Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the or is selected.

When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers disks with holes in the center. For example, consider the region bounded above by the graph of the function and below by the graph of the function over the interval When this region is revolved around the the result is a solid with a cavity in the middle, and the slices are washers.

The graph of the function and a representative washer are shown in Figure a and b. The cross-sectional area, then, is the area of the outer circle less the area of the inner circle.