39–44. Shell method about other lines Let R be the region boun...

Question

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.

y = 2

y = 2

Accepted Answer

Welcome back, everyone. In this problem, let R be the region bounded by Y equals X2, X equals 2, and Y equals 0. Find the volume of the solid formed when R is revolved above the line Y equals 5 using the shell method. Now to help us figure out the volume of the solid formed when R is revolved about the line Y equals 5, let's try to visualize exactly what's going on here. So let's do a quick sketch of our X and Y axis, OK. And we know that the region is bounded by the curve Y equals X2. So that's a parabola opening upward with a vertex at the origin. OK. So Y equals X2. We also know that it is bounded by X equals 2, so that's going to be a vertical line, OK, that cuts the X axis at the 0.2. So this is X equals 2, and by the line Y equals 0, which is the X axis. So that means our region R would be this region shaded in red on our graph, OK? So this is our region R. And this is the region that we're going to revolve about the line Y equals 5. So again, if I go back to my uh my line here, and let me put this in blue. This blue line would be the line Y equals 5. It's a horizontal line that cuts the Y axis at the 0.5. It cuts the y axis at 5. So let me represent it as this dotted blue line, OK? So the question is, how can we find the volume of the solid that's formed here using the shell method? What do we know about the shell method? Well, recall that by the shell method, the volume is going to be equal to 2 pi. Multiplied by the definite integral between A and B of or radius as a function of Y multiplied by our height as a function of Y with respect toy where D Y represents our thickness. Now, in this case, if we can figure out what A, B, R of Y, and H of Y are, then we should be able to solve for our volume using the shell method. So what do we already know? Well, A and B represent our upper and lower limits. And using the shell method with horizontal shells, since the axis Y equals 5 is horizontal, then we need to find our limits for our Y values. Notice that based on our graph or from our graph, we know that the lower limit for X would be 0 and that X equals 0, Y equals 0. On the other hand, Our region is bound by X equals 2 on the right. So at the upper limit, the highest possible value of X, which is 2, then our y value will be equal to 2 squared or 4. So this tells us that A is 0, where Y equals 0 and B would be equal to 4 because Y equals 4. Now, since we are using the shell method with horizontal shells, it would be a good idea to rewrite our equation in terms of Y. Given Y equals X squared, then that means that X is going to be equal to plus or minus root Y. Now if we think about it, our region is in the first quadrant. It's bounded in the first quadrant. So that means we can take the positive root here of X. So that means X will be equal to positive root Y. Now at a height, why, OK. Now that we have this information, then we can say that our radius R of Y, which is the distance to the axis Y equals 5, is going to be equal to 5 minus Y. On the other hand, the height, that is the length in the X direction H of Y will be equal to 2 minus root y. So in that case, this tells us that our volume V. Is going to be equal to 2 multiplied by the definite integral between 0 and 4 of 5 minus Y multiplied by 2 minus rooty with respect toy. Now we can expand and solve. So that's gonna be 2 pi multiplied by the definite integral. Of 10 minus 5 root y minus 2y plus Y to the power of three halves with respect toy and now we could rewrite root Y here as Y to the power of a half that would make it easier for us to integrate and now if we integrate here then we're gonna have 2 i multiplied by the integral of 1010 Y. Minus the integral of 5 y to the half 10/3 of y to the of 3 halves minus the integral of 2y y squared plus the integral of y to the poor of three halves, which is 2/5 of y to the poor of 5 halves and all of this is between the bones of 0 and 4. Now we can go ahead to simplify and solve, OK. So if we substitute here, this is gonna be equal to 2 multiplied by our expression with the upper limit. So that's gonna be, we're substituting 4 for Y basically. So that's 10 multiplied by 4 minus 10/3 multiplied by 4 to the power of 3 halves, minus 4 squared plus 2/5 of 4 of 5 halves. And if we substitute our lower limit, since all of our terms have Y as an expression, then our lower limit will ultimately work out to zero. No simplifying our expression. 10 multiplied by 4 is 40. Our next term, 10/3 multiplied by 4 to the core of 3 houses would be 80/3, 4 squared is 16, and the 2/5 multiplied by 4 to the core of 5 halves would be 645. Now, here, if we combine all of these in one denominator, OK, then this is gonna be 2 pi multiplied by 152/15. OK. And that would be equal to 304 pi divided by 15. Thus our volume would be 300 for a pie divided by 15 cubic units. Thanks a lot for watching everyone. I hope this video helped.

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.y = 2

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39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.

y = 2