45–48. Shell and washer methods about other lines Use both the...

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45–48. Shell and washer methods about other lines Use both the shell method and the washer method to find the volume of the solid that is generated when the region in the first quadrant bounded by y = x²,y=1, and x=0 is revolved about the following lines.

x = -1

x = -1

Accepted Answer

Welcome back, everyone. Let R be the region in the first quadrant bounded by Y equals 2 X2, Y equals 2, and X equals 0. Find the volume of the solid formed when R is revolved above the line X equals -3 using the shell method. No What are we trying to figure out here? Let's try to visualize our problem, OK? So let's draw a rough sketch of the curves that enclose the region and draw the axis of revolution. So here, let's imagine that we have our X and Y axis. And we know that this region is bounded by Y equals 2x2, which would be a parabola opening upward. So I equals 2 x 2. Next, it's bounded by Y equals 2, which would be a horizontal line cutting the y axis at 2, OK. And it's also bounded by the line X equals 0, which is the Y axis. So in this case, our area shaded in red here, OK, on our diagram, let me shade it a little bit better. But this shaded region would be our region R, OK. And now we're trying to figure out what the volume of the solid would be formed when R is revolved above the line X equals -3. So if I put that line in blue, that will be the vertical, a vertical line that cuts the X axis at -3, OK? So let me represent it as a vertical dotted line. This is where we're going to be revolving our solid. So how can we figure out, how can we figure out what the volume is by using the shell method? What do we know about the shell method? Well, recall, the shell method basically tells us that the volume is going to be equal to 2 pi multiplied by the definite integral between A and B of the radius as a function of X multiplied by the height as a function of X with respect to X, where DX represents our thickness. So if we can express our radius in terms of X or height in terms of X and find our boundaries or lower and upper limit A and B. Then we should be able to solve for the volume formed when R is revolved about our line. Now, let's first start by trying to find our limits. That means we're trying to find in the first quadrant the point of intersection of Y equals 2 X squared and Y equals 2. At the point of intersection, OK, at the point of or points of intersection, we don't know how many points of intersection there are yet, but at that 0.2 X squared will be equal to 2. If we divide both sides by 2, that means X2 will equal 1, and thus we can say X2 minus 1 will be equal to 0, which means that X + 1 multiplied by X minus 1 equals 0. Therefore, X equals 1 and X equals -1. Now remember, we are interested in the point of intersection in the first quadrant. X equals -1 is not in the first quadrant, so we're going to use our upper limit as X equals 1, OK? So, in other words, B would be 1 in our definite integra. On the other hand, we know that the region is bounded by x equals 0 at the left. So therefore, A would be equal to 0, OK, or lower limit equals 0. Thus X runs from 0 to 1. Now what about the radius and the height? Well, OK, to use the shell method, taking vertical strips at X on the interval from 0 to 1, each strip has a radius from the axis. X equals -3, OK. To say then that our radius is going to be equal to X minus -3 or X + 3. On the other hand, the height of the shell H of X will be the difference between our upper value of X minus or lower value of, of Y minus the lower value of y. And if we look at our graph, Y equals 2 is above Y equals to x2. So H of X will be equal to 2 minus 2 X2. So since we now have R X H X, A and B, then applying that to our formula for volume using the shell method, that means the volume equals 2 pi multiplied by the definite integra between 0 and 1 of X + 3 multiplied by 2 minus 2 X squad or with respect to X. Now if we expand that, it's gonna be 2 multiplied by our definite integral of 2 X + 6 minus 2X cubed minus 6 X squad or with respect to X. And now we can go ahead and integrate. Integrating each term separately, OK, then we're going to have the integral of 2 X X squad of 66 X of 2 X cubed, which is X to the fourth divided by 2, and of 6 X2d which is 2X cubed. And again these are between the bones of 0 and 1. Now we can apply our limits for each, so substituting our upper limit into our expression, that's gonna be 1 squared plus 6 multiplied by 1 minus 1 to the fourth divided by 2, minus 2 multiplied by 1 Q. And for a lower limit, if all of these terms have X, so if we substitute 0, our end result will be 0. No, if we simplify it even further. 1 squared is 16 multiplied by 1 is 61 to the 4th divided by 2, that's a half, and 2 multiplied by 1 cubed would still be 2. In this case, We can go ahead and solve here and now we are going to get uh 2 pi, OK. So that's 2 pi multiplied by 9 halves and when we simplify that, we get our volume to be equal to 9 pi. Thus, the volume of the solid is 9 pi cubic units. Thanks a lot for watching everyone. I hope this video helped.

45–48. Shell and washer methods about other lines Use both the shell method and the washer method to find the volume of the solid that is generated when the region in the first quadrant bounded by y = x²,y=1, and x=0 is revolved about the following lines. x = -1

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45–48. Shell and washer methods about other lines Use both the shell method and the washer method to find the volume of the solid that is generated when the region in the first quadrant bounded by y = x²,y=1, and x=0 is revolved about the following lines.

x = -1