64–68. Shell method Use the shell method to find the volume of...

Question

64–68. Shell method Use the shell method to find the volume of the following solids.

The solid formed when a hole of radius 3 is drilled symmetrically along the axis of a right circular cone of radius 6 and height 9

Accepted Answer

Hello. In this video, we are told that a whole of radius 2 units is drilled symmetrically along the axis of a right circular cylinder of radius 4 units and height of 8 units. By using the shell method, we want to determine its volume. So first, let's go ahead and draw a picture to represent what the information is talking about. Now, the overall, the overall shape that is having a hole drilled into it is a right circular cylinder of radius 4. So we are going to have a big right circular cylinder. And this cylinder has an overall radius of 4. And it has a total height of 8 units. Now, in the middle of the cylinder, we are drilling a hole of radius 2. So in the middle of the cylinder, we're going to drill. A symmetric section down the middle, and again, this symmetric section is going to have a radius of 2. Now, how can we use the hall method in order to determine this volume? Well, by using the shell method, the volume is going to be defined as two pi multiplied by a definite integral from A to B. Of the radius of the cylinder multiplied by its height. Now, in this case here, because the cylinder itself is right circular, we are going to integrate with respect to X. And of course, we only want to integrate over the defined boundary, which means that we don't want to integrate anything that was cut out from the shell. So what this means is that we only want the volume of the shape that exists from a distance of 2 m to a distance of 4 m, and that means that the bounds of integration are going to be from 2 to 4. Now we need to go ahead and define the overall radius of the cylinder after the hole is drilled out. Now, in this case here, we're going to allow the radius to just be some measure of X. So radius is going to be defined as X. And the height is given to us as a constant value, the height of the cylinder is going to be 8 units. So now that we have the boundaries of integration, the radius and the height, we can go ahead and construct the integral that is going to give us the volume. So volume is going to be defined as 2 pi multiplied by the integral from 2 to 4 of radius multiplied by height, which is going to be 8 XDX. And now what we are going to do is we are going to solve the integral in order to find the volume. Now, the first thing we can go ahead and do is pull the 8 in front of the integral as a constant. That is going to leave us with 16 pi multiplied by the integral from 2 to 4 of XDX. Then by taking the antiderivative of the integral, we are left with 16 pi multiplied by the quantity, 1/2 X squared, evaluated from 2 to 4. By combining the constant terms together, we will be left with 8 pi X squared, and this will be evaluated from 2 to 4. Now, by applying the fundamental theorem of calculus part two, this is going to simplify to be 8 pi multiplied by 4 quad minus 8 pi multiplied by 2 squad. By multiplying these terms together, 4 squad will simplify to be 16, and 2 squad will simplify to be 4. 8 i multiplied by 16, will give us 128 i. And -8 i multiplied by 4 will give us negative 32 pi. And combining these two terms together is going to give us a final volume of 96 pi. So this is going to be the volume of the right circular cylinder of radius 4 when we drill out a section that is 2 centimeters or 2 units in length. And with that being said, the overall solution to this problem is going to be D. So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

64–68. Shell method Use the shell method to find the volume of the following solids.The solid formed when a hole of radius 3 is drilled symmetrically along the axis of a right circular cone of radius 6 and height 9

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64–68. Shell method Use the shell method to find the volume of the following solids.

The solid formed when a hole of radius 3 is drilled symmetrically along the axis of a right circular cone of radius 6 and height 9