For the following regions R, determine which is greater—the vo...

Question

For the following regions R, determine which is greater—the volume of the solid generated when R is revolved about the x-axis or about the y-axis.

R is bounded by y=x^2 and y=√8x.

Accepted Answer

Welcome back, everyone. In this problem, let region T be bounded by Y equals X2 and Y equals 6 X, which is greater. The volume generated when T is revolved about the X axis or about the Y axis. Now, let's try to visualize what's going on here, OK? So, basically, we're saying that we have a solid, OK, that is gonna be formed because it's going to generate a volume from the equations of Y equals X2, which would look something like this, and Y equals 6 X which would look something like this, OK? But no, we're saying that we want to figure out where would we get a greater volume if we revive it? Would we get a greater volume if we revive it about the X axis or if we revolve it about the y axis? That's what we're trying to figure out. So here notice that we can use the washer method to find the volumes because we'll have an outer and inner radius in the solid that's formed. So let's try to find the volume when it's rotated about the X axis and when it's revolved about the Y axis and then compare each. So let's start with when T is revealed about the X-axis, OK. When T is revived about. The x axis Now what do we know about revolutions about the x axis? Well, recall that the volume is going to be equal to the definite integral between the bonds of A and B of pi multiplied by Y22 minus Y12 with respect to X, OK, where Y2 is the outer radius and Y1 is the inner radius, so Y2 is greater than or equal to Y1. And both of these must be positive on or interval from A to B. So what we'll need to do here, let me just draw a line to separate both sides of our problem. But what we'll need to do here essentially is that we need to first find the intersection points so that we can find our limits A and B, then figure out which of these equations, whether Y equals X2 and Y equals 6 X is Y2 and Y1, and then we can plug them into our formula for the volume and solve. So let's first start by finding the intersection points. So finding the intersection points. We can equate both equations. In other words, The intersection points are at the point where X squad is equal to 6 X. Well, in that case, that means X2 minus 6 X will be equal to 0, which means X multiplied by X minus 6 equals 0, and that implies that either X equals 0 and X equals 6. Since X equals 0 is less, then that would be our lower limit A and X equals 6 would be or upper limit B, OK? Now that we have our limits, we can figure out which one of the equations would be Y2 and which would be Y1. How can we find that? Well, we can do that by substituting a value into each equation and seeing which is greater. So let's take uh let's set X to 1. If we let X equals 1, then Y equals X2 would be equal to 12, which is 1, and Y equals 6 X equals 6 multiplied by 1, which is 6. Therefore, since Y equals X2 is less than Y equals 6 X, Y equals X2 would be Y1 and Y equals 6 X would be Y2. So if we plug all of this information into our formula, then that means or volume when rotated or revolved above the x axis would be equal to pi multiplied by the definite integra between 0 and 6 of 6 X squared. Minus x 2 squad all with respect to x. If we expand what's in our bracket, that's gonna be multiplied by the definite integral between 0 and 6 of 36 X squad minus X to the fourth with respect to X, and now we can integrate both of these separately because here the integral of 36 X squared is going to be 12 X cubed while the integral of X to the 4th will be X 5th divided by 5. Now that's between the bones of 0 and 6. Now if we think about it here, for our lower limit, if we substitute 0, it will, or the result, sorry, will be 0, OK? So that's, that would end up to be zero. So let's go ahead and sub for X equals 6. So that's pi multiplied by 12 multiplied by 6 cubed. Minus 6 to 5th divided by 5. I know if we solve here, then we should get this to be equal to 1,036.8 cubic units. So this is the volume when our solid is when T is revealed about the X axis. Now, let's try to figure it out when T is revolved about the Y axis. So we can go back to our problem, OK? And let me just move our little sketch here. And let's work this side in blue, OK? So now on the right we're going to figure out what happens or the volume when T is revealed about the y axis. Now, how do we use the washer method for the revolution about the y axis? Well, no, that volume v of Y will be equal to the definite integral between A and B of pi multiplied by X2 square minus X1 squared with respect toy. Where X2 and X1 are functions of Y and X2 represents the outer radius while X1 represents the inner radius. So X2 is greater than or equal to X1, which is greater than or equal to 0 on our interval from A to B. And now remember this interval is referring to the Y axis. Before our intervals for the X axis, no, this one is for the Y axis. So that means we'll need to change our limits, OK, A and B, and we'll, again, we'll also need to figure out which of our equations would be X2 and which would be X1. But first, we'd have to express both of our equations in terms of Y. We need to make X the subject. So let's first start with the limits. No recall before our limits were 0 and 6. When X equals 0, Y equals X2 would be Y equals 02 or 0. So this would be our limit A. Or lower limit. On the other hand, when X equals 6, Y equals 6 squad, or 6 multiplied by 6, which is 36, so our upper limit B is going to be 36. So now we have our limits in terms of why. Now, before we find X1 and X2, let's express our equations in terms of Y. So, for Y equals X2, OK. Then X, let's say if, if, rather, if Y equals X2. X would be equal to root Y, but if Y equals 6 X, then X would be equal toy divided by 6. So if we let y be equal to 1. Then X would be equal to root 1, which equals 1 for X equals root Y and for X equals Y divided by 6, X would be equal to 16. So since uh X equals root Y is greater. OK, then we can say that X2. Is equal to root Y. And we can say X1 equals Y divided by 6, OK? So in that case, that means our volume is going to be equal to pi multiplied by the integral between 0 and 36. Of Roy squared. Minus Y divided by 62 with respect toy. If we continue here, so that's gonna be pi multiplied by the integral between 0 and 36 of Y minus Y squared divided by 36 with respect to Y, OK. And no. If we integrate our terms here, this is gonna be pi multiplied by Y2 divided by 2 minus Y cubed. Divided by 108, OK, between the bounds of 0 and 36. And again if we substitute 0 into or for our lower limit that will be 0. So really this is going to be pi multiplied by 36 2 divided by 2 minus 36 cubed divided by 108. And now when we go ahead and simplify, we get our volume when it's rotated above the y axis to be equal to 216 pi cubic units. No, what do, what, what does this tell us? Well, by comparison, here, we can tell that VX, OK, is greater than VY. As a matter of fact, I should probably move it to put it in between here. So VX is greater than VY, OK. And since VX is greater than VY, that means that. The volumes, the volume about the X axis is greater. Thanks a lot for watching everyone. I hope this video helped.

For the following regions R, determine which is greater—the volume of the solid generated when R is revolved about the x-axis or about the y-axis.

R is bounded by y=x^2 and y=√8x.

Key Concepts

Finding the Region of Integration

Volume of Solids of Revolution Using the Disk/Washer Method

Comparing Volumes Generated by Different Axes of Revolution

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