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In this problem, we are using integrals to understand area.
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And specifically, this chapter in your textbook is dealing with things called trigonometric substitution.
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And we're not going to do that exact process in this problem, but we are going to be using inverse trigidrigg functions.
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And you'll see that this comes up later in this problem.
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So the main point of this problem is we have this curve given for why.
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And we're given a boundary for a x plus y.
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Now, what does this mean? we're going to be dealing with some kind of circle, and we'll be understanding how we can geometrically interpret that in order to calculate or evaluate intervals.
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We're given y equals x squared over 2.
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We're also given that x squared plus y squared is less than equal to 8.
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And this here should look like a very similar structure of an expression, that is the expression of a circle with a radius of the square root of 8.
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So what we can say is that we see this intercept in our y space.
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So we can say y equals the positive portion of the square root of 8 minus x squared.
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And again, our radius of this circle is a square root of 8.
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So what does that mean? the area of this circle is 8 pi.
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But what what we need to do is we need to find where exactly our curve intersects this circle, because we're not finding the area of the entire circle.
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That's already given to us by the definition of the circle or the circle equation.
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So what can we do? we can say that we can rearrange y to get 2y equals x squared.
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Now if we compare this with the other equation, we can say y is less than equal to negative of 4 or 2.
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So what does that mean? our points of intersection are the points 2 -2 and negative 2 -2 -2.
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So when y is less than or equal to 2, x is less than equal to 2 or negative 2.
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Now, we have to use integration to find the area of our different regions, specifically the cut region defined by our curve and then the leftover region...