Solve the given problems by integration. Find the volume...

Solve the given problems by integration.
Find the volume generated when the first-quadrant area bounded by $y=e^{x}$ and $x=2$ is rotated about the $x$ -axis. [Hint: After setting up the integral, rewrite. $(e^{x})^{2} \text { as } (e^{x})(e^{x}).$]

Key Concepts

Disk (or Washer) Method

The Disk Method is a technique for finding volumes of solids of revolution. It involves slicing the solid perpendicular to the axis of rotation and approximating each slice as a disk (or washer, if there is a hole). The volume is then expressed as the integral of the area of these slices along the axis of revolution.

Setting Up the Integral Expression

Formulating the integral requires correctly expressing the radius of each disk in terms of the variable of integration, and then determining the appropriate limits of integration based on the boundaries of the original region. This process often involves rewriting the integrand to facilitate simpler integration.

Integration Limits

Determining the correct limits of integration is key to accurately modeling the region being rotated. The limits are set based on the intersection points or boundaries of the region with respect to the chosen variable, ensuring that the entire region is accounted for in the volume calculation.

Integration of Exponential Functions

Integrating exponential functions, such as those in the form e^(kx), is a common task in calculus. The antiderivative of e^(kx) is (1/k)e^(kx), which is essential when evaluating integrals that arise in problems involving exponential curves.

Transcript

00:01 In this question, we want to find the volume generated by this curve and bounded by x equals to 2 and rotated about the x -sacist.

00:10 In general, when we have a curve, y equals to fx, and we want to find the volume from x -equals to a to b generated about the x -sacist over here.

00:24 We'll just get a small strip over here.

00:30 The width is so small, we'll just call it dx.

00:32 And of course the height here is y.

00:40 Now as it generates over the excesses, it's creating a circular disk.

00:47 The circular disk, the width is dx or the height of the disk is dx, and the radius is y.

00:58 So the volume of this disk is pi r square h, right? pi y square dx.

01:08 To get the entire.

01:09 Volume from a to b, we will just sum up.

01:13 Now since this is a continuous summation, so it will be an integration sign from a to b, pi y square dx.

01:22 You can leave the pi outside, and so that will be your volume.

01:31 So to find a volume for this, so volume is equal to pi summing from 0 to 2 y square dx so that will be e to the power of x times e to power x because that's y square d x there are a couple of ways to integrate this but for this topic we're using substitution method and integration power rule so for substitution method let's u be e to power x the u over the x is e to power x so d u is e to power x d x.

02:17 So this part over here is d u.

02:20 Now we need to substitute these x values here as well with u values.

02:26 So when x equals to the lower limit 0, u is e to power 0 and that is 1.

02:35 When x is the upper limit of 2, u is e square...

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