Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord! Numerade Educator ### Problem 5 Medium Difficulty # Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.$$y=x^{3}, y=x, x \geqslant 0 ; \quad \text { about the } x-axis$$ ### Answer ## Volume of the solid formed is$\frac{4 \pi}{21}\$

#### Topics

Applications of Integration

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##### Top Calculus 2 / BC Educators ##### Heather Z.

Oregon State University  ##### Michael J.

Idaho State University ### Video Transcript

I want to find the volume of the solid. If my equations are white, was X cubed y equals x X greater than or equal to zero. And I'm rotating about the X axis. So first of all, I want to draw these graphs. Why equals X? Cubed is the cubic graph cubic function and that graph I know we wouldn't try and zoom this in a little bit. That graph looks like this and then for y equals X. I know that I'm looking at a linear function that goes through the origin and has a slope of one so linear function that goes through the origin slope of one. Try my best to draw a straight line here. Thanks. So let's pretend that's actually straight line. That means the area that's greater than zero bounded. Here is this party just shaded in green. So we're revolving around the X axis. Notice how from the party shaded in green to the X X is that we're revolving around. There's a lot of empty space, so that means this is gonna be a washer for the washer set up going around the X axis. We want to use pi times the in a girl capital R squared minus lower case R squared DX and will use x one and X two as the bounds Capital R squared is like if it were a disk and there was nothing missing, lower case R is the stuff that we're trying to take away. It's the volume we're removing from this figure so that if we have, you know, this solid volume and then we want this empty space in the middle, the lower case, ours this part we're deleting away. So with that in mind, let's first find our ex boundaries. I think those are gonna be the easiest year. Pretty obvious that both these functions go through the origin. And then if I set them equal to each other to solve, then I would subtract to get them all in one side. I would factor out the X they have in common, which gives me that zero and then the X squared minus one. Give me plus or minus one. Since I'm only looking at the positive values of X based on the original set up, I'm only gonna use X equals one. Here is my other boundary. So for the integral. We're gonna go from 0 to 1 now. We needed a hot A filet and capital r in lower case R. I think it's easiest to look for the stuff we wouldn't take away first lower case R. So lower case R is what I'm removing. The radius I'm removing. So it has to start from, you know, wherever you're revolving around the access and then it goes up until it touches the area. So for that distance upper minus lower, it's touching the cubic graph going down to the access. So, for lower case R, I would fill in x cute for capital are. It's like What if I were looking at this and there was no space missing at all? Like if it was a perfect disc problem. So if it went all the way from the access all the way up to the top, can you see that all the way from the access all the way up to the top? So if it's that capital, our access to the top, then it's actually gonna be this linear function minus the access. It has to go all the way up to that straight line and come all the way down to the axis. If I had wanted it to be filled in, Okay, that's what it takes to be the limb. So that capital are. Then we have here as X minus zero, the upper function minus the lower function, which obviously we can just write his ex. The squares are part of your set up, so don't change that. We just filled in R and R, and at this point, it's pretty easy to just look term by term for this anti derivative. We have X squared, and this isn't X to the sixth. So for those anti derivatives, the 1/3 X cubed minus 1/7 extra the seventh and then going from 0 to 1. So if I feel in my upper bound minus the lower boned up around his wine, that lower round is zero, and that zero stuff is actually just gonna drop off. Then for my final answer, I would write high times 1/3 minus 1/7 or if I want to look really nice and go and put those together, I could say four pi over 21 getting common denominators there and combining them together Oklahoma Baptist University

#### Topics

Applications of Integration

##### Top Calculus 2 / BC Educators ##### Heather Z.

Oregon State University  ##### Michael J.

Idaho State University 