Solid of revolution Compute the volume of the solid of revolut...

Question

Solid of revolution Compute the volume of the solid of revolution that results when the region in Exercise 85 is revolved about the x-axis.

Accepted Answer

Welcome back everyone. In this problem, we want to determine the volume of the solid of revolution that results when the region bounded by X2 + Y2 equals 4, X equals 2, and Y equals 2 of X divided by 2 is revolved above the x axis. And here we have a graph of our curve and the circle. Now how can we figure out the volume of the solid of revolution about the x axis? What do we know? Well, recall that we can use the washer method, and the washer method basically tells us that the volume is going to be equal to pi multiplied by the integral between the lower and upper boundary of Y2 quad minus Y12 with respect to X where Y2 is the upper curve and Y1 is the lower curve. If we can identify these unknowns from our equation, then we should be able to find our volume. So let's see if we can do that. Now, based on the given graph, we can tell that the lower boundary is X equals 0 and the upper boundary is X equals 2. So that means then that A in our integral will be 0 and B will be equal to 2. Next, from our graph, we can also tell that the upper curve is Y equals 2 shake of X divided by 2. So we can call that Y2. Which must mean that our lower curve is is from the equation of the circle X2 + y2 equals 4. But now remember we are integrating with respect to X, so we'll need to rewrite our problems solving for Y, expressing our equation in terms of X. Now if X2 + Y2 equals 4. Then if we subtract X2d from both sides and find the square root, then Y in this case or a lower curve Y1 will be equal to the square root of 4 minus X2. Now that we have our bones and our upper and lower curves, then we can substitute it to solve for our volume. Because no this would mean then that the volume is going to be equal to pi multiplied by the integral between 0 and 2 of 2 of x divided by 2 squad, which is going to be 4 squad of x divided by 2. Minus the square of the square root of 4 minus X2, the square cancels the square root, and that would simply leave us with 4 minus X2, and we are integrating all of that with respect to X. Now if we take out our uh our frack our. Brackets here, or parentheses, sorry, this is going to become -4, OK, plus X2, because if we subtract the negative X2, it's going to become positive X2. So we're finding the integral of 4 squad of X divided by 2 minus 4 + X2. How can we do that? Well, let's try to find the integral of each term. That's how we can solve, OK? So we're gonna have pi outside as our constant still, but now for a first term we're going to rewrite it as 4 multiplied by 2 multiplied by the integral of check squared of x divided by 2, OK, with respect to x divided by 2. Next, we're going to find the integral of 4 with respect to X. And then we'll also find the integral of X squared with respect to X. So now let's go ahead and find the integral of each term. No for our Shake square term, it's integral eventually will work out to 8an of X divided by 2. That's after we find the integral and multiplied by 4 multiplied by 2, which is it. Next, the integral of 4 with respect to X is 4 X, and the interval of X squared with respect to X is going to be a third of XQ. So this is our integral and all we need to do now is to apply our limits. So in this case, for our upper limit, it's going to be 8 of 2 divided by 2 or 1. So that's 8 multiplied by the son of 1 minus 4 multiplied by 2, OK, which is 8. Let me just write it here. Plus 2 cubed, which is 8 divided by 3. So that's what happens when we substitute our upper limit 2. No for or lower limit 8 multiplied by the than of 0 would be 04 multiplied by 0 is 0, and 0 cubed divided by 3 is 0. So all of those in our second limit would eventually solve the 0. And know that we have this, then we can go ahead and try to solve because this is going to be pi multiplied by 80 of 1 minus 16/3 because negative 8. Plus 8/3 equals -163. And if we factor it out from both of these terms, then we'll end up with 8 pi multiplied by the sum of 1 minus 2/3. Thus, the volume of our region when it is revolved above the x-axis is 8 pi multiplied by 1 minus 2/3 cubic units. Thanks a lot for watching everyone. I hope this video helped.

Solid of revolution Compute the volume of the solid of revolution that results when the region in Exercise 85 is revolved about the x-axis.

Key Concepts

Solid of Revolution

Disk and Washer Methods

Definite Integrals for Volume Calculation

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