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 -3 using the shell method. Now, in order for us to figure out the volume of the solid formed when R is revolved, let's try to visualize exactly what our problem is telling us. So here, let me put our X and Y axis. OK. And from our problem, we know that we have the curve Y equals X2d. That's a parabola, that that that's opening upward and its vertex is at the origin, OK. Next, we're told that we have the line X equals 2, which will be a vertical line that cuts the X axis at 2. And finally, we're told that it's bounded by Y equals 0, which is the X axis. So, on our diagram, our region R would be this region shaded in red. OK. So let's call this region R. And no, we're trying to find the volume of the solid formed when R is revolt, the line Y equals -3. So the line Y equals -3 will be a horizontal line that cuts the Y axis at -3. So let me represent it as this dotted line on our graph, OK? So how are we going to figure out the volume of the solid formed when R is revived using the shell method? What do we know about the shell method? Well, recall that the shell method basically tells us the volume will be equal to 2 pi multiplied by the definite integral between the lower and upper limits of the radius as a function of Y multiplied by the height as a function of Y multiplied by the thickness of our solid D Y. So If we can figure out what R of Y and H of Y are and also N B, then we should be able to integrate to figure out our volume. What do we already know? Well, we know that at X equals 0, Y equals 0, and that's important because X equals 0 is is our lower limit if it were to be or if we were finding the volume with respect to X. So Y equals 0 would correspond to that. On the other hand, we know that our region stops at X equals 2, and when X equals 2, Y equals 2 squad, which is 4. So this tells us that our lower bound. A is 0 and our upper bound B is 4. Now what else do we know? Well, remember we're trying to find V in terms of Y. And we know that Y equals X squared, so we can express X in terms of Y, which is going to be plus or minus root Y if we square both sides. And now remember, uh, we are going to, we are working with the first quadrant or region is in the first quadrant. So we are going to take the positive root. So, so X equals root Y, OK? Now, if X equals root Y. Then that can help us to figure out our radius and our height because at a height Y, OK, we were using the shell method with the horizontal shells since the axis Y equals -3 is horizontal, then at a height of Y or radius R of Y will be equal to Y minus -3, which is equal to Y + 3. And that's because it's the perpendicular distance to the axis Y equals -3. On the other hand, the height as a function of Y is going to be the length in the X direction, which is going to be 2 minus root Y. So now that we have A, B, R of Y, and H of Y, we can integrate the soul for volume because no, this means our volume V is equal to 2 pi multiplied by the definite integral between uh between 0 and 4. Of Y + 3 multiplied by 2 minus root y all with respect to y. Now if we expand that, that's going to be 2 multiplied by our definite integral of 2y plus 6 minus y to the power of 3 halves, minus 3 Y to the power of 1/2 or with respect toy. And now we can integrate by integrating each term separately. The integral of 2 Y is Y squared of 6 is 6Y. of Y to the poor of three halves is going to be 2/5 of Y to 5 halves, and of 3 Y12 it's going to be 2 Y to the power of 3 halves, and that's between the limits of 0 and 4. Now if we apply our limits substituting 4 in our upper in our expression for our upper limit, that's gonna be 4 squared plus 6 multiplied by 4 minus 2/5 of 4 to the power of 5 halves minus 2 multiplied by 4 to the power of 3 halves. And on the other hand, if we substitute 0 into our expression for the lower limit, we know that that's just going to simplify to 0. Now, if we solve, so that's gonna be 2 pi multiplied by 16 plus 24 minus 645, OK, minus 16. And when we combine these, OK, then we're gonna have 2 i multiplied by 56 divided by 5. And that is going to be equal to 112 pi divided by 5. Thus the volume is going to be 112 divided by 5 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