Let R be the region bounded by the following curves. Find the ...

Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=|x| and y=2−x^2; about the x-axis

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says let TP the region bounded by Y is equal to the absolute value of X and Y is equal to 6 minus X2. What is the volume of the solid formed when T is revolved about the y axis? And below the question we're given a graph that has our two functions plotted on it. So it has Y is equal to 6 minus X2 and Y is equal to the absolute value of X plotted on it. We also have our region T identified on this graph, so it is a shaded region between these two functions, where the shaded region is bounded above by the function Y is equal to 6 minus X2, and is bounded below by Y is equal to the absolute value of X. So, the first thing we need to do is determine the intersection points between our two curves. And if we look at a graph, we see that that should occur approximately around X equal to + and minus 2. But let's find that algebraically. So, for the first case, we're gonna be looking at the case when X is negative. So that means that Y is equal to the absolute value of X can be rewritten as Y is equal to minus X, where X is less than 0. And so that means we'll just need to set these two y values equal to each other. So we'll have minus X is equal to 6 minus X squad. And so we'll just need to solve this for X. So the first step is to add X to both sides of this equation. So we'll get 0 is equal to 6 plus x minus X squad. Now this quantity can be factored, so that means that 0 is going to be equal to the quantity of 3 minus X in quantity, multiplied by the quantity of 2 + X in quantity. So now we just need to set each factor equal to 0 and solve for x. So for the first factor we'll have 3 minus X is equal to 0. So that means that X is going to be equal to 3. And for the second factor, we will have 2 plus x is equal to 0. So that means that X is going to be equal to minus 2. Now for this, we assume that X had to be negative. So that means that X equal to 3 is not a valid solution. So the only solution that we have is X is equal to -2. Now, we're going to repeat this process for the intersection point that occurs when X is positive. So that means that Y equal to the absolute value of X is just going to be Y is equal to X. So again, setting the two Y values equal to each other will have that X is equal to 6 minus X2. And again, we need to solve for X, so we'll subtract X from both sides of the equation. So we'll have 0 is equal to 6 minus X minus X squared. And again, this could be factored, so we'll have 0 is equal to the quantity of 3 plus x. In quantity multiplied by the quantity of 2 minus x in quantity, and again we need to set each factor equal to 0 and solve for x. So for the first factor we'll have that 3 plus x is equal to 0. So that means that X is equal to minus 3, and for the second factor we'll have 2 minus X is equal to 0, and therefore X is going to be equal to 2. And here we assume that X had to be a positive value. So that means that X equal to -3 is not a valid solution. So the only solution that we have here is X equals to 2. So that means that our intersection points between these two curves occur at X equal to plus and minus 2. So, now we can set up the integral for our volume. So, we call that our volume V. Is going to be equal to. Pi multiplied by the integral from 2 to 2 of the quantity, and here we need to take the curve that bounds our region from above, which in our case is going to be the quantity of 6 minus X squared and quantity and we'll need to square that curve. Then we'll have to do minus, and here we're going to subtract off the square of the curve below. So the curve that bounds our region from below is going to be the absolute value of X, and then that quantity needs to be squared in quantity, DX. So it's going to be integrating with respect to X. So now we just need to simplify our integrand. So we'll just multiply everything out, so our volume V is going to be equal to. Pi multiplied by the interval from -2 to 2 of the quantity of 36. Minus 12 x squared. Plus 14th. Minus X squared in quantity d X. So here on the last term, we drop the absolute value sign, since when we square a quantity, it's always going to be a positive value, so we can drop the absolute value signs. So now we just need to collect light terms. Doing so, this is going to be equal to pi, multiplied by the interval from -2 to 2. Of the quantity of 36. Minus 13 X2. Plus x 4th in quantity. DX And so now we just need to perform the integration, and we'll just integrate term by term. In doing so, our volume V is going to be equal to pi, multiplied by the quantity, and here we integrate 36 with respect to X, we get 36 X. Then we'll have minus 13, and this gets multiplied by the integral of X2 DX. And so we recall that when we integrate X2 with spec X, we get X cubed divided by 3. Then we'll have plus, and here for the last one, we're going to be integrating X to the 4th with respect to X, which will give us X to the 5th, divided by 5 in quantity, and this needs to be evaluated from -2 to 2. So, substituting in the limits of integration, we see that this is going to be equal to. Pi multiplied by the quantity of 36, multiplied by 2. Minus 13 divided by 3 multiplied by 2 cut. Plus 2 to 5th divided by 5. Minus 36 multiplied by -2. Plus 13 divided by 3. Multiplied by -2 cute. 2 to 5th. Divided by 5. And quantity And so when we evaluate this expression, we see that this is going to be equal to 1,312 pi divided by 15. Cubic units. So this is the answer that we are looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.y=|x| and y=2−x^2; about the x-axis

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Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=|x| and y=2−x^2; about the x-axis