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=√sin x,y=1, and x=0; about the x-axis

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says lets be the region in the first quadrant bounded by the curves Y is equal to the square root of cosine of X, Y is equal to 1, and X equal to pi divided by 2. Find the volume of the solid generated when S is revolved about the X axis. And below the problem we're given a graph, and on this graph we have our three curves plotted on it. So we have the curves Y is equal to 1, X equal to pi divided by 2, and Y is equal to the square root of cosine of X plotted on this graph. We also have our region S shaded on this graph, and the region S is the region that is enclosed by these three curves. And this region is bounded above by Y is equal to 1. It's bounded on the right by X equal to pi divided by 2, and it's bounded below by Y is equal to the square root of cosine of X. So, the first step in solving this problem is to determine the X value for the intersection point between the curves, Y is equal to the square root of cosine of X and Y is equal to 1. And if we look at a graph, we see that this should occur around X equal to 0, but let's find this algebraically. So we'll set the square root. Of cosine of X. Equal to 1. So to solve for X we'll square both sides, and so that means we'll have cosine of X is equal to 1. Now, we call that cosine of X is equal to 1 at integer multiples of 2 pi. But in the problem we're we're told that we need to look in just the first quadrant. That means we're bounded on the left by X equal to 0, and our region S is bounded on the right by X is equal to pi divided by 2. So the only value of X that satisfies this equation that's in that interval is going to be X equal to 0, which agrees with what we see on the graph. So now, we have all the information that we need to find our volume, and to find the volume, we're going to use the washer method. So, our volume V. is going to be equal to, and here we call your wash method, so it's gonna be equal to pi. Multiplied by the integral, and here we're going to be integrating from 0 to high divided by 2. Of the quantity And here for the wash method, we recall that for the first term of our integram, we need to square the curve that is bound in our region from above. And so the the curve that bounds our region from above is Y is equal to 1. So we'll have 1 squared, then we'll have minus the square of the curve that bounds our region from below, and the curve that bounds our region from below is the square root of cosine of X. So we'll have minus the quantity of the square root of cosine of X. In quantity squared, in quantity DX. This is going to be integrating with respect to X. So first we'll simplify our integrand. So this is going to be equal to pi, multiplied by the integral from 0 to pi divided by 2. Of the quantity of 1 minus cosine of X in quantity DX. So now we just need to perform the integration, we can just integrate term by term. That means our volume is going to be equal to pi, multiplied by the quantity. Here for the first term, we're going to be integrating 1 with respect to X, which will give us X. Then we'll have minus for the second term, we're going to be integrating cosine of X with respect to X, which is going to give us sine of X. In quantity, and this needs to be evaluated from 0 to high divided by 2. So substituting in our limits of integration, we see that our volume is going to be equal to pi, multiplied by the quantity of pi divided by 2, minus sign of pi divided by 2. Minus 0 plus sign of 0 in quantity and simplifying this expression, this is going to be equal to pi multiplied by the quantity of pi divided by 2 minus 1 minus 0 + 0 in quantity. And evaluating this expression, we see that our volume is going to be equal to. Pie squared divided by 2. Minus high. Cubic units. But this here 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=√sin x,y=1, and x=0; 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=√sin x,y=1, and x=0; about the x-axis