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Problem

Each integral represents the volume of a solid. D…

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Problem 40 Easy Difficulty

Each integral represents the volume of a solid. Describe the solid.

$ \pi \displaystyle \int_{-1}^1 (1 -y^2)^2 dy $


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01:13

WZ

Wen Zheng

00:23

Amrita Bhasin

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 2

Volumes

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Applications of Integration

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Video Transcript

All right, let's do this problem. So the question is asking you you have the integral right here. Describe what it iss like. It's a needy question. Right. Um, it takes a lot of practice and experience for anyone to be able to really be able to tell what this shape is. So I am going to give you the hint off how you solve these kinds of problems. And there are many ways to do it, of course, but it's just a idea that would probably help you how to do it. Okay, Now, the approach that I like to take is that I see that there is a pie in something squared. Okay, so that should remind you pi r squared. So a pi r squared is basically the area of a circle. So that's the biggest hint that you have here. Okay? Furthermore, when you have a d. Y, you basically have a very, very small difference of wise telling you that it probably came from pi r squared Delta y, which is actually a thin, thin disc with radius R that looks like this. Okay, and then you're integrating it from negative 1 to 1, which is one of the process off integrations. So you're probably going to get a shape that looks like this where you cut through it. It looks like a circle, and then each off these thickness is D y. And these coordinates starts from negative 1 to 1. Okay, so that's the basic big picture off what's going on here. And if you remember, um, this is actually one of the formula that you use in calculating a shape revolved around an axis. So, for example, if you have a graph that looks like this and then if you revolve around it and access, you can get a shape that kind of looks like Okay, so this is what's really happening. The input is a Y. And truly speaking, it really doesn't matter. But I think the book wants you to think about it this way. So I am actually going to go through it one minus. Why squared? If I said X is equal to one minus y squared X is a function off. Why? Why is not a function of X, so if you graph it, it's actually going to be a sideways parabola that looks like this negative 1 to 1. It goes through one, and it's a proble that looks like this. Okay, so the distance from here to there given a value of why it is going to be one minus y squared, and this is going to be treated like the radius that I mentioned right here. So when you revolve around the Y axis, the solid that you're going to get is going to look like this. You will have a radius with respect and why, which is equal to one minus y squared. This is going to be a change in the UAE values because it's a very thin disc, and then you're going to take the integration or sum it up from negative 12 positive one. So if you add up all of the thin discs, it would look like I would describe it something like a lemon shape or a a rugby ball like shape where the radius is precisely one minus y squared or something like this. Okay, so I know my drawing is not the best, but that's exactly what we're trying to describe in this integral, and that's how you solve these problems

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