 In this video, we're gonna find the area enclosed by the ellipse x squared over a squared plus y squared over b squared equals one. Now, an ellipse is itself essentially a generalization of a circle, right? So if we take our standard x and y axes, maybe something like this, a ellipse has two radii attached to it. There's a horizontal radius, a horizontal radius, which we'll call a. This is the number that's below the x squared there, well, the square root is. And so we have a and negative a right there. Then we have this vertical radius b, which go above and below the x-axis at a mount. And these values could be the same, they're likely gonna be different. And if we kind of connect these dots in a smooth curved way, we get the basic idea of an ellipse. And that's not a perfect picture, but it's kind of like someone's squished a circle a little bit. An ellipse is a genuine circle if a and b are the same number, which would be the common uniform radius all around this thing. So how does one find the area of an ellipse? Well, first of all, because of the symmetry of the ellipse, it does turn out that we only have to find the upper semi ellipse. If we can find that, then again, despite the crudeness of my drawing right here, the symmetry is the top semi ellipse would be equivalent to the, has the same area as the bottom one. So we're just going to, we're gonna integrate here, but we're gonna just double the top there. Now, the reason why we wanna focus on the top half is that your standard ellipse fails the vertical line test. But if we only look at the top half, then it is a genuine function in our usual sense. So if you solve for y, you would subtract x squared over a squared from both sides. This gives us y squared over b squared equals one minus x squared over a squared times both sides by b squared, b there or b squared, right? This gives us y squared is equal to b squared minus b squared over a squared, x squared. And then take the square root, we're gonna get y equals, normally there of course be two square roots plus or minus the square root of b squared minus b squared over a squared, x squared, like so. But as we're taking just the upper semi, semi ellipse, we're gonna take the positive one right there. Now to make this thing a little bit cleaner, I'm actually gonna factor out b squared over a squared inside the square root here that leaves behind one minus, I'm sorry, not one, a squared minus x squared and this would sit inside the square root. The advantage of that is of course the b squared and the a squared are a perfect square and so we can simplify and take the function y equals b over a times the square root of a squared minus x squared. Now, exporter rules don't allow us to simplify the square root as much as tempted as we are to say, oh a is square is the perfect square and x squared is the perfect square and then inside square roots, we wanna simplify it. That's of course not a justified simplification whatsoever. So this right here is the function we want to integrate because we wanna find the area of the ellipse. What we're gonna do is we need to find the area under this curve doubled. So the area we're looking for capital A will be two times the integral of y dx right here. But of course, we just found a formula for y so we'll plug that in there. We get b over a times the square root of a squared minus x squared dx. Now we have to think of what are the bounds of this thing gonna be. If we're not certain, we can take a look at the picture again, right? The bounds will seem to be your cross sections would look something like this. They can come all the way over to negative a up to positive a. And so that's what we get here, negative a and positive a. Now again, using the symmetry of the ellipse, the left hand side is actually equal to the right hand side and so we notice we have this symmetric interval a to a and this is in fact an even function. So yet one more use of symmetry here, we actually get four times b over a the integral from zero to a of the square root of a squared minus x squared dx. And so this is the integral we wanna calculate in order to find the area of this said ellipse. All right, so far that was just setting up an area type problem here. What we're gonna do then is now we have to calculate the anti-derivative of this thing, the square root of a squared minus x squared. And this square root, the square root of a difference of squares, this makes me think that we probably need to do some type of trigonometric substitution. We need to eat a trig sub. And you might think that because this lecture is part of section 7.3 about trigonometric substitution, good intuition, but really we're thinking of this codex, this ancient codex we saw before, right? The square root, the square root of a squared minus x squared, this is what tells us we need to do some type of substitution here. And the substitution is gonna be x equals a sine theta. So using that substitution here, let's look at all of the ingredients. So we need to set x equal to a sine theta. And a here of course is just a constant, we'll just leave it alone. And then so trust in the codex here, we set x equal to a sine theta. And so dx when we take the derivative is gonna be a cosine theta d theta. And considering the right triangle associated to theta here, let's draw the picture. If we think of this angle here as our theta, then from this statement right here, we get that sine theta is x over a. So that is the opposite side is x. The hypotenuse is a. And if we carry through the Pythagorean equation, we plug in x and a for the opposite hypotenuse. We're gonna get that the adjacent side is the square root of a squared minus x squared. All right, and so that's useful for us because if we combine these two sides together, we get adjacent over hypotenuse, that's our cosine, cosine theta equals the square root of a squared minus x squared all over a. We're gonna prefer to clear the denominators. And so what we actually get is that the square root of a squared minus x squared equals a times cosine theta, right? I kind of was really went through that calculation very quickly there. It's very similar to what we saw in the previous example and this will be very much similar in all of these trigonometric substitutions. One often goes from here and to here pretty quickly with these trigonometric forms because they show up very similar each and every time. Now we have to make sure we plug these things in appropriately so dx will be substituted as an a cosine theta d theta and then the square root will become also an a cosine theta, all right? And so making those substitutions, I still wanna see the problem there, we're gonna end up with this four b over a integral. The square root, like I said, is gonna become an a cosine but the dx also becomes an a cosine plus also a theta. Now notice in this one, we do have a definite integral, right? We're trying to find the area from zero to a here. So what we could do is we could continue to calculate this anti-derivative, switch it back to terms of x and then plug in zero and a but as we're trying to really just find the area, we don't need a specific anti-derivative, I'm gonna switch the limits of this integral into terms of theta as well. And so we have a t chart here, we know what x is, we know what theta, well, we need to figure out theta is. So x will go from zero to a and using this identity right here, oops, I squiggle all over it, using this identity right here, we can figure out what x is, which would imply what a is. So for example, when x equals zero, you're gonna get zero equals a sine theta, divide both sides by a, you're gonna get zero equals sine. When does sine equals zero? Well, going back to our domain issue, because again, sine actually equals zero a lot, but inside this domain, right, sine would only be zero at the angle zero itself. So this tells us that when x is zero, theta is equal to zero. On the other hand though, when x equals a, we get a sine theta, divide both sides by a, we get one is equal to sine and in that domain, sine will equal one exactly at pi halves. And so that's gonna be our new bounds. We're gonna go from zero to pi halves and the convenience of switching the bounds is that we don't have to switch back to x ever in this problem, we can proceed forward with just theta. All right, and so then what can we do here? We have a four b over a, there's two a's inside of the integrand that goes as an a square, which we can factor out. We're gonna get an a squared right there, integrate from zero to pi halves. We could cosine squared theta d theta. I can simplify the a's a little bit, one a on the bottom cancels with one a on the top. And now how do we deal with the cosine squared? Integrating that, it can be somewhat of a challenge and you wanna use that half angle identity that we've seen before. Specifically, cosine squared theta is equal to one half one plus cosine of two theta. Make that substitution in right here. So that would give us a four a b integrate from zero to pi halves. And then we get this one half one plus cosine of two theta d theta. Notice that the one half on the inside would combine with the four on the outside giving us a two. We're gonna integrate the one and the cosine of two theta separately. So we have a two a b in front. The antiderivative of one is theta. The antiderivative of cosine of two theta will be one half sine of two theta as we go from zero to pi halves. And so what happens is we plug these things in here to a b. So we plug in the pi halves, we're gonna get pi halves plus one half sine of pi the two pi over two becomes a pi. Notice, I'm just gonna mention this already. Sine of pi is equal to zero. So that's gonna disappear. But then we have to subtract from this when we plug into zero. We plug in zero for theta, that's a zero. And then we have to also subtract one half sine of zero. And as we already mentioned earlier, sine of zero is zero. So everything is gonna disappear except for this pi halves for which when you multiply that by the two a b, that simplifies to give us our final result, which is gonna be pi a b like so. So we've now derived an area formula for an ellipse. The ellipse is area is just gonna be pi times its horizontal radius and its vertical radius. And compare this to a standard circle. Remember, a circle is a uniform ellipse where the horizontal and vertical radii are the same. So in that situation, we're saying that a equals b and we'll call them r. Well, if you plug those in right here, you're gonna get that the area is equal to pi rr, that is pi r squared. The standard equation of the area of a circle, which we probably already know and love. And so this integral does in fact generalize the area formula of a circle, but it does require a trigonometric substitution in order to help us get that formula correctly here.