Area and volume The region R is bounded by the curves x = y²+2...

Question

Area and volume The region R is bounded by the curves x = y²+2,y=x−4, and y=0 (see figure).

b. Write a single integral that gives the volume of the solid generated when R is revolved about the x-axis.

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Let S be the region bounded by the curves, X equals Y 2 + 3, X equals Y + 5, and Y equals 0. Write a single integral that gives the volume of the solid generated when S is revolved about the y axis. Fantastic. So it appears for this particular problem we're asked to determine. What the single interval that gives the volume of the solid that is generated when this regions is revolved about the Y axis. So we're trying to figure out what this interval is that gives the volume of the solid generated by S when it's revolved around the Y axis, and that is our final answer that we're ultimately trying to solve for what is this interval? So now that we know that we're trying to solve for this integral value based on all the information that is provided to us by the prom itself, let's take a moment to investigate our graph that is provided to us by the prom itself. We have our Y axis, which is our vertical axis, which goes from a negative region going up to 4, and then we have our X axis, which goes from what appears to be about -1 to 8. And then we have our three curves where our blue curve represents Y equals 0. Our black curve represents X is equal to Y2 + 3, and our red curve is represented by X is equal to Y + 5. And the region S is highlighted in green. Fantastic. And it appears to be shaped like half a heart, almost, or maybe almost like half a teardrop. Shape is for our shaded region in green for S. So, now that we have a visual of what's happening for this particular prom. Let us begin to solve for our final answer. So our first step that we need to take is we need to find the intersection points of X is equal to Y2 + 3 and X is equal to Y + 5. Fantastic. So with that in mind, In order to do that, we need to set 2 + 3 is equal to Y + 5 to help us get the following. Expression to the power 2 minus Y minus 2 is equal to 0, and all we did is we set these two expressions equal to each other and then we rearranged it to, so we would have it equal to 0 on one side. So now we need to for our second step is we need to solve for Y2 minus Y minus 2 is equal to 0, and when we do that, we will find that Y to the power 2 or Y2 minus 2y plus Y minus 2 is equal to 0. And we now need to factor out our expression. So when we begin the factor, we will find that Y multiplied by parentheses Y minus 2 + 1 multiplied by parentheses Y minus 2 is equal to 0. And we will note that Parentheses Y minus 2 multiplied by parentheses Y plus 1 will be equal to 0. So that will mean that undoubtedly that Y is equal to 2. And Y is equal to -1, fantastic. Our third step now is we need to note that since Y is equal to 0 is a boundary, we need to integrate from Y is equal to 0 to Y is equal to 2. Fantastic. So that will mean that for our 4th step, we need to determine that our outer radius, which we'll call OR, our outer radius which will denote as our subscript O, so outer radius, will be equal to X, which is equal to Y + 5, and our inner radius, which we'll call R subscript I, R I will be equal to X, which will be equal to Y to the power of 2. +3, fantastic. And our 5th step now is we need to determine the volume integral, and we will recall a note that V is equal to the integral evaluated from 0 to 2 of pi multiplied by square brackets of. Parentheses of Y + 52 minus parentheses Y2 + 32 D Y. And our 6th and final step is that we need to note that since the region is bounded by Y is equal to 0 and Y is equal to 2. The correct answer for our integral, the final answer that we're ultimately trying to solve for will be the integral from 0 to 2 of pi multiplied by square brackets of parentheses Y + 52 minus Y2 + 32. Don't forget your square bracket DY. And that is our final answer. Hooray, we did it. So looking at the problem itself, to recap what we've learned, that will mean that our final answer, without a doubt, will have to be the following. So once again, our final answer undoubtedly will be the integral evaluated from 0 to 2 of pi multiplied by square brackets of parentheses Y + 52 minus Y2 + 32 D Y and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Area and volume The region R is bounded by the curves x = y²+2,y=x−4, and y=0 (see figure).
b. Write a single integral that gives the volume of the solid generated when R is revolved about the x-axis.

Key Concepts

Region Bounded by Curves

Volume of Solids of Revolution

Setting up a Single Integral for Volume

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