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The parabola $ y = \frac{1}{2} x^2 $ divides the disk $ x^2 + y^2 \le 8 $ into two parts. Find the areas of both parts.

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

Integration Techniques

Campbell University

Harvey Mudd College

University of Nottingham

Boston College

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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the proble y equals X squared over two divides the disc X square plus y square less than equal today into two parts. Let's find the area of these parts s o N red. We have our circle. This is the boundary of the disc, and Red Export plus y squared equals eight. So it's a circle with a radius radically. And then it's intersected by our problem at thes two points highlighted agreed. So there's a blue problem like equals X squared over two. So a one let's do Nobody won the area inside the circle So inside of the red but a bluff above the blue problem and then a two is everything else in a circle Or if you want everything in the circle but below the problem. So let's find a one first. So here will find a one. Then use the fact that a one plus a two Well, if you add a one and a two together, you're getting the area of the entire circle so we know the area of a circle. Oh pi r squared. We're here. We have our squared equals eight. So this equals a pie and then we have softer A too Wei have this so really, all of our work is going to be on finding a one. And then once we find it, we find a to using this formula circled and then we're done so is focused on a one first. So here, this looks like in a roll with respect to the X from this point of intersection and green to the other one. And then we have top minus bottom. So this is the area between two curves. So notice that the talkers, which is part of the circle X squared plus y squared, equals eight. If you saw for why you get to parts of the circle the plus radical, which is the top half. And then on the bottom half of the circle down here, below the X axis, you have the negative radical. So that's our tankers. So we're doing top man in his bottom, and then the bottom curve is down here on blue. That's X squared over two. Now we need the points of intersection. So let's go on to the next page to do that. Well, we're used The fact that why was explored over too, and then we'LL plug it into this equation over here, so I'm gonna put equals a because at the point of intersection, run the on the circle, not inside the disc. And so the radius of that circle isthe root. Wait, So this is the equation of the circle now playing in excluding him too, for why we obtained this equation. So let's go ahead and simplify this Get X to the force over four plus X squared minus a equals zero. And then let's go ahead and multiply both sides by four. And this is a contract IQ, but not an expert in X square. And this one will factor as X squared plus c and then X squared minus four. Now, if you see this first term, X squared plus eight is always positive. It's always bigger than zero because X squared is bigger than zero. So, really, this term here is bigger than our equal to eight, which is bigger than zero. So this first term is number zero and then we set this term zero, and we get X equals plus or minus two. So these are the X values of the points of intersection. So we have negative tune into top man in his bottom And now, to make this a little easier I'LL go ahead and basically using the cemetery on the original area A want so recall a one was of this form So a one is the collection these two regions right here and basically by cemetery instead of finding it from negative to to to we'LL just find it from zero to and then we'll supply by two so we can just do this instead, Not a necessary step. Might make it a little easier. And now, because you could see this first, Integral has a radical. We should use the tricks up here for the second one. We could just use the hour rule. So let's just go ahead and split this into two animals. So writing that all we have minus and then we cancel the twos Explorer dx, Sierra two and we would just evaluate these separately, put them together, and then we'LL get our anyone so looking at this easier integral. Here. You can just do Tyrol. You're excused over three zero to, and then we get eight over three after plugging in the end points so that just leaves us with this remaining in a girl over here and we'LL do it tricks up for this Let's go to the next page. This was our inaugural. So here this is a trick sub Let's do X equals radically signed data And since this is a definite no rule, we might be best off here If we go out and simple by this radical we'LL see why shortly it could help us avoid writing the triangle. So here this is R X, go ahead and take the X And now they will see why this was a good choice because now we could try to find the new limits of immigration in terms of data. So plugging in X equals zero into this equation and also using the fact that when you do a tricks of involving sign, you have to make sure it is between negative power to in parenting. So let's use these fast tto find data. So on the left side for X, we have zero and the right side to room two Scient Ada, this means sign has to be zero, and sense data is in this interval Over here. We must have data equals zero is the only solution into sensible now for the upper limit. Tune said the left hand side X equal to two and on the right two room to sign Data Sol for sign. And this is a point in the circle and this interval, the point that we want this state a equals power for So that's our new upper limit lower and let's do one more step here we have eight minus X squared inside the radical that becomes radical a minus and then X squared is eight signed square. We could fatten throughout a radical one minus sign square. And then if we like we can write this radical is to room two use with Agron identity and then finally just cancelled a square with the square root. Okay, so that takes care of the Radicals. Let's go ahead and plug this all into our our new interval. In terms of data, we have to our limits change. Now we have zero pi over four. We just simplify the radical that was to room two cosign data and then the ex. I was too rude to cause and data self here we have radical eight times radical eight. So we have eight and then we multiply by this two on the outside, we could probably sixteen and then we have co signed Square. So trying to remember what to do in this case and then for this trick in a rural there's no science present. So in this case, the best way to go is one plus coastline to date over. So let's pull that too out. We have sixteen over to since eight zero power before one plus co sign to data, and this we can integrate. We have data plus sign to Theatre Over to and points Sierra Power before Dan. Let's go to the next patient, plugging those end points and simplify so plugging in pi over four for data. But and then when we plug in zero, we have zero plus sign zero over, too, so we could ignore the second term sign. A pie or two is one. So let's go ahead and multiply. That ate through and we get right over here and then we have to buy and then times in an eight times a half, it's to my place for so remember that this was just the first integral We also had to subtract eight over three from this from the previous integral that we got using the powerful. So let me take a step back. A one was equal to This is the animal that we just completed using the tricks up. Originally, we had this other expression that we used. So this number corresponds to this first in the room. So we have to pi plus four. And then we're subtracting the previous answer the quadratic and we had eight over three for this one. So put these fractions together. Forest well over three. So we subtract eh for over three. So this is a one and then recall for a tu es tu is everything else in the circle. The formula that we had for a two was a pi minus a one. A pi was the area of the whole circle. A one is over here on the right, So plugging this in we have a pie minus to Piper's for over three. So we have six by minus four over three and there's a final answer

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