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 \ge 0 $ ; about the x-axis

Volume of the solid formed is $\frac{4 \pi}{21}$

Applications of Integration

You must be signed in to discuss.

Numerade Educator

Missouri State University

University of Nottingham

Idaho State University

So in this problem, we're finding the volume of the solid obtained by rotating the region founded by a Y equals execute. Wyffels X with X is greater than zero. So we can first start off by drawing on the reading. I'm going first. You'll notice that I put the majority of the area in the excess positive. Gary and Wyatt was positive area. And the reason I did this is because we know that excess positive and when X is positive. Why is also positive? Because pies with Cuba is always positive, as is a pilot raised two to the power. So the majority of the action will be going fun going to happen in the first clock ins. We can first draw the y equals X line that was like that. And if you think about it, y equals acts y equals execute those two flying slash curves intersect at the point at one one. They also intersect at the point zero zero because both both of those points satisfy the equations of the line. Zero Q zero and also zero equals zero. Once cured, this one in one equals one. So where you draw out the second line white because ex cued, this point will be one one, and this point will be zero zero. And we're rotating this about the X axis. So let me first off by drying start by highlighting the region that we're rotating and red. We're rotating this about the X axis, which means that we are forming a washer, which is made out of subtracting the smaller disk out of a larger disc. So here is the radius. So as you can imagine, there's a smaller disc. Here's the larger disk. We only care about this region between the disks. Thus the reason why we call it a washer. So to do this when you know once again find the cross sectional area on in this time because we're rotating about the X axis when you find it in terms of X, because as exchanges, the area changes as well, because the washer this time it's a circle subtracted from another circle. So this is equal to pi r squared and you can make the bigger circle of big R. The smaller ship will smile, So as you can see here, the y equals X line, which is the top line right here is creating the bigger circle. And then the Y equals X to the Q X to the three line or a curve is creating the smaller circle because it is less than the Y equals X line below it, pretty much the entire in a roll from zero to one, and that makes sense when you keep a fraction, you make it smaller. So the are right here represents the y equals excellence We can put that in. So our is just why, in this case, a just plugging X and then the small R is just why equals excuse plug in skewed and that is equal to pi. I'm pulling out the pie here. So the X squared minus X to the six. So that is our cross sectional area function in terms of X, so we can go ahead and do our anymore. Now, as I said before, we're integrating from X equals zero two X equals one and we put in the function that we just found. So I'm going to immediately pull up. I hear because pie has no relation with the ex, Then you can put an X squared minus x for the six d x because that is what we found before to evaluate this interval when you find the anti derivative of X squared in excess of six. Oh, that is just X cubed over three minus X to the seven first seven Valentin from zero to one. We don't really care about the zero case because that was his Give us zeros. So we only care about the one case. And when we do that, we get pie. I went there minus one sex. And if you do a little bit more math, you'll finally arrive at the answer, which is four pi over twenty one.