(a) Calculate the area between y = sin(2x), the lines x =...

(a) Calculate the area between $y = \sin(2x)$, the lines $x = \frac{\pi}{4}$ and $x = -\frac{\pi}{4}$, and the $x$-axis. (b) Find the volume of the solid of revolution created when the area bounded by $f(x) = 2e^{-x^2}$ (shown in the diagram), the $x$-axis and the lines $x = 0$ and $x = 1$ is rotated around the $y$-axis. (c) The density function of a thin rod lying along the $x$-axis is given by $\delta(x) = \begin{cases} -2x + 1 & -1 \le x < 0\\ 2 & 0 \le x \le 1 \end{cases}$ Find the mass of the rod.

Transcript

00:01 Okay, for this question, we are given a region bounded by two parabolas.

00:05 I have them marked on this graph here.

00:07 We have y equals x squared in red and then 6x minus x squared in green.

00:11 So we're going to revolve it around the x -axis and the y -axis.

00:16 So let's do the x -axis first.

00:18 And because this problem is asking just for a setup of the integrals, that's all we'll do in the solution.

00:24 And then from there, it's requesting to use a calculator to get the value.

00:28 So let's see how we would set these up.

00:31 For an x -axis revolution, i would probably end up going with a disk method or a washer method for this.

00:39 The reason why is because the radii are easily pulled from the graph.

00:46 So you'll notice that we'll have a small radius and a large radius.

00:55 So this will be our capital r and little r.

00:59 The way you find the radii is you go from the axis of revolution to the point.

01:04 Closest boundary to get your little r and then to the farthest boundary to get your largest r.

01:12 So in this case, your volume integral using the washer technique is going to be pi on the outside and then r squared minus r squared.

01:25 And in particular, let's let's find out what these radius functions should be.

01:32 Capital r is the height of our green curve, which means it will be 6x minus x squared, whereas our little r is the height of our red parabola, which is x squared.

01:50 Because these functions are in terms of x, that tells us that our integral is going to be in terms of x as well.

01:58 So what does our volume look like? we have pi integral 6x minus x squared, quantity squared.

02:09 Be careful.

02:10 Remember in the expression, each of the radii are squared.

02:14 And then minus x squared, squared, which would be x to the fourth.

02:20 So you have your integral setup.

02:22 The only thing that's left to do is to find the bounds.

02:25 And how do we find the bounds? well, first of all, from the graph that i have drawn, you can see that one of them is certainly going to be zero.

02:33 And you don't have to take that for granted.

02:35 We can actually set them equal to one another and solve...

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