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Sketch the region enclosed by the given curves and find its area.

$ x = 2y^2 $ , $ x = 4 + y^2 $

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Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

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University of Michigan - Ann Arbor

Lectures

01:01

Sketch the region enclosed…

05:00

04:47

02:23

04:06

0:00

03:20

03:08

All right, let's sketch the region cost by those 2 curve, serves first, so here you saw our x y plane and now x, equals to 2 equals to 2 y square. So it's a problem open to the right. It looks like this. This is our x goes through 2 y square and the second 1. Also a problem is some x equals to 4 plus y square, so is also open to the open to the right. So here is 4, since the coefficient is 1 instead of 2, which means this first curve you'll catch up with the second curve. So you have the intersection right. So the intersection intersect here- and this is symmetric- the x axis- all right. So eventually we'll get this. Some moon, ship, enclosed area and you observe that everything is represented by y. It'S like is our independent variable here. So to find this area we take the integral integral with respect to y okay. So then we need to find a boundary for y and as we see this, i call this y 1. This is our y 2, so this is our y 1. This is our y of the intersection point. So how do we find this 1 way to? Basically, we just plug in those 2 functions. You know this intersection happens when 2 y square equals to 4 plus y square. So we can solve this. We know this is y equals to. We move this y square to the other side, so we will get 1 y squared equals to 4. So we get y equals to positive or negative 2 point. So since our 1 is negative to 1 equal to negative 2 and our 2 will go to positive point, so our integral goes from negative 2 to 2 point and the thing inside the integral is just the right curve: minus the left curve. So our right curve, where is 4 plus y square minus our last curve, our left curve, is some left curve, is 2 y square okay. So basically, this is just 4 minus y square and anti derivative for 4 minus y squared 4 y minus 1 cube 1. Third y cube and you mietheboundary y equals to 2 and y e goes to negative 2. So this goes to the first. 1 is 8 minus 8 over 3 minus active 8 plus 8 over 3 point right. So this as to 32 over 3.

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