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Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places.

$$y=x^{2}, x^{2}+y^{2}=1, y \geqslant 0$$

(a) About the $x$ -axis (b) About the $y$ -axis

$$

V \approx 1

$$

Applications of Integration

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So if I want to find value, then I need to know the area and rotate it. So first, Well, my crocs here x squared and then actually, the circle equation X squared plus y squared equals one. So the area bounded by these two equations. Is this part here? I'm shaking. Now. I need to know where I'm revolving around in this problem. Let's revolving around the X X is so if we're roaming around the X axis. Noticed all this missing piece here from the volume, which means it's a washer set up now because it's the washer set up. We know that volume is pi times the integral R squared minus r squared. We use DX because it's the X axis, which means I also need X for my limits of the integral. So x one and X two are gonna be the bounds there. Now, the other thing that's important or helpful is I noticed that this has symmetry. So personally, I would like to just use this portion of the shaded area and then double my set up. So I'm actually going Teoh be solving two pi. I'm gonna go from zero to wherever they meet here, so I still need to find this X is my ending limit. And then still the same stuff R squared minus R squared DX. It just makes it easier especially. But I am working out my hand to use a zero there for the lower bound on the limit. And this is a time when it's great to use a calculator because we're gonna get to decimals here but again, on opportunity to make it simpler work by using that zero bound. Now, if I want to see what this looks like in the graphing calculator I would want to solve for Y, and solving for y equals you would have y equals plus or minus one minus Sorry, square root of one minus X squared, right? You subtract the X squared over first, and then you square root both sides and because you square, where did you get? Plus or minus? The negative only defines this lower part of the circle, so I actually don't need the lower part of the circle for my volume. So really, it's just the positive version of the square root. So as I'm graphing it, I'm just going to see this top curve which didn't see the top part of the circle, which is good, because that's what I need. I would recommend using graphing calculator to find this intersection point so I would graph X squared, and I would grab the square root of one minus X squared to find where those two graphs are equal to find that X to value. So if I use the intersection of my graph to solve this, I see that X is let me tell you exactly what I got. 0.786151 I may go to several decimal places, so that way, at the end, I can actually still list my answer to five decimal places. So if I want my answer to be five decimal places, I would like for my work to go to at least six. Help me keep it exact. So, feeling it, this is my upper bound on the integral. So it goes from zero toe 00.786151 Now I need to find capital R squared and lower case R squared. So capital R squared is what if this volume were all the way filled in? Nothing was missing. That would mean I would shade everything going from the top curve down to my access of revolution. Well, the tub curve is defined by the circle. The circle is the square root of one minus X squared, and then access of revolution is your X axis. So that's just a minus zero. Not super helpful as I'm solving through that part of it, right? Doesn't really define anything there, so I don't have to write the minus. Seagram, lower case R is what are we taking away? Well, we're taking away all this missing stuff below the shaded area until we get to the axis. So below the shaded area, notice is defined by your parabola down to the access. So actually X squared down to zero. Okay, X squared down to zero. I could just right as X squared. Now, the in time this integral is really awesome to rewrite because you have a square root that's gonna cancel out. And then you just have that X squared squared, which becomes an extra the fourth so pretty easy to rewrite the integral part of us. But then maybe not a simple Teoh evaluate that crazy decimal. So you could go one step further to get this anti derivative. Or at this point, you could type this whole thing into the calculator. So if I were to take that whole thing into my calculator multiplying by two Pi and Brian the Integral I get 3.544 5912345 decimal places. That is a decimal approximation of my volume here for the graphs between that X squared in the circle function going around the X axis.

Oklahoma Baptist University

Applications of Integration