Let R be the region bounded by the curve y=cos^−1x and the x-a...

Question

Let R be the region bounded by the curve y=cos^−1x and the x-axis on [0, 1]. A solid of revolution is obtained by revolving R about the y-axis (see figures).

b. Find an expression for the area A(y) of a cross section of the solid at a point y in [0,π/2].

Accepted Answer

Welcome back everyone. In this problem, a solid of revolution is formed by revolving a region bound by the curve Y equals the axion of X and the X axis on the interval 01 above the y axis. What is the expression for the area B of Y of a cross section of the solid at a point Y in the interval from 0 to 1/2 of pi. A says B of Y equals pis squared Y. B square, sorry, CY squared, and D pi multiplied by 1 minus sysquared. Now how are we going to figure out the correct expression for the area BFY? Well, let's try to think about what we're visualizing here, OK? So, we are looking for a region that's bounded by the curve Y equals the axion of X, which would probably look something like this, and the X-axis, OK, that's revolving about the Y axis. OK? So, and that's between the bones or the bones of 0 and 1. So if I were to put it on a graph here, it would probably look something like this. This is our region that we're going to revolve about the Y axis, OK? So, how can we find an expression for the area BF Y? Well, for starters, we'll need to express the boundary curve in terms of X. And since Y is equal to the arc sign of X. Then this means X would be equal to the sign of Y. No, the cross section at a height Y is a disk of radius X equals sin Y, OK? And remember that since a ring is going to be formed or a disk rather is going to be formed, that means the cross section is going to be circular, which means that the area of the cross section B of Y. Is going to be equal to pi multiplied by r squared where our radius is siny so that's pi multiplied by the sin of y squared which we could rewrite as pi sine squared y. Now if we evaluate the bones for this in terms of why, OK, when X equals 0. Then Y will be equal to the arc sign of 0, which is 0. And when X equals 1, Y equals the arc sign of 1, which equals 1/2 of pi. So that tells us the expression B of Y equals pi squared Y is for Y in the interval from 0 to 1/2 of pi, which fits the information in our problem statement. Therefore, that tells us that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Let R be the region bounded by the curve y=cos^−1x and the x-axis on [0, 1]. A solid of revolution is obtained by revolving R about the y-axis (see figures).

b. Find an expression for the area A(y) of a cross section of the solid at a point y in [0,π/2].

Key Concepts

Inverse Trigonometric Functions

Cross-Sectional Area of a Solid of Revolution

Relationship Between x and y in the Region

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