9-34. Shell method Let R be the region bounded by the followin...

Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

{Use of Tech} y² = ln x,y² = ln x³, and y=2; about the x-axis

Accepted Answer

Welcome back, everyone. In this problem, let us be the region bounded by Y2 E equals LNX. Y2 E equals LNX 5th, and Y E equals 1. Find the volume of the solid generated when S is revolved above the X-axis using the shell method. And from our graph here, we can say C or region bounded. S. Now what do we know about the shell method that can help us to figure out the volume of our solid? Well, recall that by the shell method. The volume is going to be equal to 2 pie. Multiplied by the integral from A to B, that is the limits on or y axis. Of the radius, OK, as a function of Y multiplied by the height as a function of y with respect toy. So if we can find our limits or radius and our height, then we should be able to solve for the volume of the solid generated when S is revolved above the X axis. So let's first start with our limits. We're taking horizontal shells since we are revolving about the X axis. Now from our graph here, we can see that Y or region for S is going to be on the interval. From 0 to 1, so our lower bound is 0 and our upper bound is 1. And because we are taking horizontal shells for this given way, then the radius of the shell are would be equal toy, OK, and the height of our shell. Is going to be equal to the XX as a function of Y to the right, minus X as a function of Y to the left. OK. And now when we look at our graph, we can tell that Y2 E equals LN X is to the right. And if we rewrite that, in other words, making X the subject of our equation, then we would have it as E to the power of Y2. While our function to the left would be Y2 equals L and X 5th, which again, if we rewrite that in terms of X, it would be Epo of Y2 divided by 5. OK? So this is our height. So since we have the radius, the height, and our bones, then we can plug it into our volume or formula for the volume, because no, this implies that the volume is going to be equal to 2 pi multiplied by the definite integral between 0 and 1 of our radius Y multiplied by our height e to the power of Y2 minus e to the Y 2 divided by 5, all with respect to Y. Now, if we distribute our Y term here, this will be too pie. Multiplied by or integral between 0 and 1 of ye to the y squared with respect to y. Minus or integral of ye to the y square divided by 5 with respect to y. And we wrote it this way because now we can integrate both of these using substitution. So now integrating both functions. Let's first start with YE to the Y squared B Y. Now for this function, OK, as we said, we're using substitution. So let's let you be equal to Y2. This means that the the derivative of U with respect to Y would be equal to 2 Y. and if we do some rearranging here, then Y D Y would be equal to a half of the u. Now remember, we'll also have to rewrite our bones in terms of you. And when Y equals 0, then you equals 0. While when Y equals 1, then you equals 1. So this implies then that our derivative in terms of Y. Would be equal to the derivative between 0 and 1 or integral in terms of why my apologies would be equal to the integral between 0 and 1 of e to the 4 of u multiplied by dU divide by 2 in terms of u. Now we know we can take a half oath as a constant of our integral, and if we integrate e to the poor of you with respect to you, then its integral is e to the poor of you and remember it's still between the bones of 0 and 1. Now if we substitute our bones to help us solve, we're going to have e to the power of 1, which is E minus e to the power of 0, which is 1. So the integral is a half of E minus 1. Next, let's do the same and try to find the integral by substitution of Ye to the power of Y squared divided by 5 with respect to Y. Now, in this case, we can let T, let's use a different uh variable for substitution. T equal Y2 divided by 5. In that case, This would mean that the derivative of T with respect to Y would be equal to 2 Y divided by 5, which means that YDY is going to be equal to 5 halves of DT. On the other hand, changing our bones, when Y equals 0, then T equals 0. Well, if Y equals 1, then T equals 12 divided by 5 or 1/5. In that case, then this means that are integral in terms of why. It is going to be equal to the integral between 0 and 1/5 of E to the pore of T multiplied by 5 halves DT, OK, in terms of T. And again we can take our constant 05 halves and thus we're finding the integral of e to the power of T with respect to T, which equals e to the power of T between the bones of 0 and 15th. And when we substitute our bones, that's going to be e to the power of 1/5 minus e to the power of 0, which is 1. So now that we've found the integral for both of our terms separately, we can combine them back into our equation for the volume. Because no, this means the volume then will be equal to 2 pie. Multiplied by 1/2 of E minus 1. Minus 5 halves of e to the poor of 1/5 minus 1. And now notice that we can factor to half from both terms, OK, and multiply it by 2 from the 2 pi to cancer and that is going to leave us. With pi as our constant on the side and on the inside we'll have E minus 1 minus 5 multiplied by E to the 15 minus 1, which would be -5 E to the 1/5 plus 5. We can simplify that even further because -1 + 5 is 4. So our volume is going to be pi multiplied by E + 4 minus 5 E to the 1/5 cubic units. This is the volume of the solid generated when S is revolved above the X-axis using the shell method. Thanks a lot for watching everyone. I hope this video helped.

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis. {Use of Tech} y² = ln x,y² = ln x³, and y=2; about the x-axis

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9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

{Use of Tech} y² = ln x,y² = ln x³, and y=2; about the x-axis