Let R be the region bounded by the following curves. Find the ...

Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=0,y=lnx,y=2, and x=0; about the y-axis

Accepted Answer

Welcome back, everyone. In this problem, let T be the region bounded by Y equals 0, Y equals LN X, Y equals 1, and X equals 0. Find the volume of the solid generated when T is revolved above the Y axis. And here we have a diagram showing our region T. Now, if we're going to figure out the volume of the solid generated when T is revolved about the Y axis, then it would be a good idea for us to express the region in terms of Y. We know that it lies between the y axis X equals 0, between Y equals 0, that is the X axis between Y equals 1 and the curve Y equals NN X. So let's express or curve Y equals LN X in terms of Y. OK? In other words, let's solve for X. And by definition X will be equal to e to the pore of Y based on the definition of the natural log. Now for our region, we also know that it's bond between Y equals 0 and Y equals 1, OK? So we can write that or represent our boundaries here. Now in this case, how can we find the volume of the sallid generated? Well, let's use the disk method. By the disk method. OK. Then we know the volume is going to be equal to the integral between its limits A and B of the area of the area in terms of Y with respect to Y, OK? And now in this case. Because we're using the disk method revolving around the y axis and the slices are horizontal at a height of Y, OK, then the radius as a function of Y will be equal to E to the power of Y, which means that the area as a function of Y, since it's a disk, will be pi multiplied by R of Y, which is E to the power of Y squared. And this gives us pi e to the power of 2 Y. So with that information. We can evaluate our integra. Because no, this must mean that our volume is going to be the integral between 0 and 1 of pi E to the 2 Y with respect to Y. I know if we take out our constant i from the integral that's pi multiplied by the integral of e to the power of 2y with respect to y. When we integrate it to the power of 2 Y. It leaves us with a half of e to the power of 2y between our bones of 0 and 1. You know if we substitute our bones, that will be pi multiplied by a half of e squared minus 1/2 of e to the of 0. If we think about it, we can factor out a half, so that gives us uh a half of pie and no. Inside our bracket we're left with e squared minus e to the power of 0, which is 1. Thus, our volume when the solid or the volume of the solid generated when T is revolved above the y axis will be equal to a half of pi multiplied by Ed minus 1 cubic units. Thanks a lot for watching everyone. I hope this video helped.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.y=0,y=lnx,y=2, and x=0; about the y-axis

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Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.
y=0,y=lnx,y=2, and x=0; about the y-axis