[MUSIC] Now that we've introduced the idea of an area function, we can apply this idea to find some perhaps surprising one expected connections. The first is to explain the relationship that we noted in earlier videos between the formulae between the perimeter and area of a circle. Suppose we have a circle of radius r. Consider the effects on area and perimeter that occur by making a small change in the radius. Say Î”r, which is positive in this diagram. Color the area of the original circle in blue, and color the pathway around the circle with Î”r in beige. So the blue area we can call A, which is a function or r, given by the formula pi r squared. The beige area is the change in A, Î”A, which is the difference between A evaluated at (r + Î”r), and A evaluated at r. Let P denote the perimeter of the small circle, this area is shaded in blue. Also a function of r, given by the formula 2 pi r. The area of the beige pathway is represented symbolically by Î”A, the resultant change in the red is by Î”r, and the pathway has two sides. A smaller side length, which is the perimeter of the circle with red as r, and a larger side length, which is the perimeter of the circle with radius r + Î”r. Such area must be bounded below by the small perimeter and multiplied by the width of the path. And bounded above by the larger perimeter multiplied by the same path width. And so we get this chain of inequalities. We can divide through by Î”r,, and get to Î”A divided by Î”r, sandwiched between P(r) and P(r + Î”r). As Î”r approaches 0, P(r) just stays the same. And P(r + Î”r) becomes P(r), because the perimeter function is continuous. And Î”A divided by Î”r becomes derivative dA/dr. All three limits must coincide by the squeeze law because of the chain of inequalities, and the fact that the limits on the left and right are equal. We conclude that the derivative dA/dr becomes the perimeter function P, explaining this fact that we've noted before directly from the respective formulae. It's natural then to ask if there's a similar connection between a volume of a sphere and its surface area. Consider a sphere of radius r. Denote its volume by V and its surface area by S, both of which are functions of r. Consider again, now in the context of the sphere, the effects of a small positive change, Î”r, in the radius. This will propagate small changes in both the volume and surface area of the sphere. The change in volume Î”V, will be the volume of the outer layer, which we will refer to as the crust, by analogy with the crust of the earth. This crust, so to speak, is colored light beige, and the outer surface of the expanded sphere is colored slightly darker beige. The inner sphere, the core, with original radius r, is colored blue. The crust has uniform thickness of Î”r throughout. Where it interfaces with the blue inner sphere, we have a smaller blue surface area. So the volume of the crust must be bounded below by the blue surface area, whatever that might be, multiplied by Î”r. On the other hand, the crust interfaces with the exterior outside world through the largest surface area of the expanded sphere. So the volume of the crust must be bounded above by the darker beige surface area multiplied by Î”r. And so we get a chain of inequalities. But the surface area of the inner blue core is represented symbolically by S(r). And the darker beige, surface area of the larger sphere by s(r) + Î”r. So we can write our chain of inequalities more concisely. Dividing through by Î”r, we get s(r) is less than or equal to Î”v / Î”r which is less than or equal s(r + Î”r). As Î”r approaches 0, S(r) of course, stays the same. S(r) + Î”r goes to S(r) by continuity of the surface area function, and Î”v/Î”r becomes the derivative dv/dr. By the squeeze law, again, all three limits are equal. This proves the that the derivative of the volume of the sphere, must be the surface area. Whatever the formula is for the volume, we can get a formula for the surface area by differentiating. We've established this amazing connection, simply by thought experiments and a few diagrams. To apply the fruit of our labors, we'd like to come up with some actual formulae. With some lateral thinking, we can figure out a formula for the volume fairly quickly. And then use this indirectly to get the surface area by differentiating. It's actually very difficult to find the surface area formula directly. Here's a sphere of radius r, again, but imagine a copy of the xy plane forming a vertical cross-section. So that we get this circular profile that crosses the axis at plus and minus r. The circle that we see in this profile has equation xÂ² + yÂ² = rÂ². So in the xy plain, y takes the value Â± âˆšrÂ²- xÂ². Now comes the lateral thinking, we imagine slicing through the sphere vertically, but in a direction perpendicular to the xy plain. To produce this green circular vertical cross-section with the x axis passing exactly through its center. If this green circle intersects the x axis at a particular value of x, then its radius is the positive âˆšrÂ²- xÂ². We want the total volume of the sphere, and the green cross-section is like an infinitely thin slice. The idea intuitively, is to imagine the area of all this infinitely thin, vertical cross-sections being added together, so to speak, as x moves from minus r to plus r, along the x-axis. It's like the continuous sums that we've been discussing for the definite integral. But now, we use the cross-section area of these circles as the integrand. Here are just a few snapshots of this cross-section as x moves from left to right. But in fact, we're thinking of a continuous movement from x equals minus r through to x equals r. To set up the integration, we need to know the area of the circlular cross-section for any particular x. But the radius is y=âˆšrÂ²- xÂ², and the area of the circle is Ï€yÂ², which equals Ï€(rÂ²-xÂ²). So we're really looking at the definite integral with integrand Ï€(rÂ²-xÂ²). In terminals Â±r, as we're thinking of x as moving all along the interval from -r to +r. In setting this up, we're appealing to your intuition. We're in fact using something called the disk method taught in more advanced courses in calculus. Which formally relaxed it's definite integral to the limit of Riemann's sums where the pieces being added up volumes of so called disks with certain thicknesses. Which become in the limit, as things get thinner and thinner, out infinitely thin circular cross-sections. The volume we're finding is also called a volume of revolution, which you can look up and read about if you wish. So the volume of this sphere is this particular definite integral, and the constant Ï€ can come out the front. We wish to form an antiderivative of rÂ²-xÂ², and evaluate it for x between -r and +r. To find this antiderivative, observe that rÂ² is a constant, so that contributes rÂ²*x. Then we take away an antiderivative of xÂ², which is xÂ³/3. These pieces combine to give us our full antiderivative. And evaluating it for x between -r and r, simplifies to 4/3 Ï€rÂ³. Thus we get the well-known formula for the volume of a sphere. We now get an explicit formula for the surface area by taking the derivative with respect to r, which quickly simplifies to 4Ï€rÂ². Thus finally, we reproduce the well known formula for the surface area of a sphere. In today's video, we explained in detail the connection between areas and derivatives, which is the main idea that leads to a proof of the Fundamental Theorem of Calculus. To do this, we changed that point of view, and considered an area function built out of a definite integral, where the upper terminal is a variable. The derivative of this function turns out to be the value of the integrand. Using properties of anti-derivatives, the statement of the fundamental theorem quickly followed. We then applied the main idea involving an area function to explain why the derivative of the area with respect to the radius produces the perimeter. We adapted that idea to explain why the derivative of the volume of a sphere with respect to the radius produces the surface area. We then worked out a formula for the volume of a sphere directly, by integrating the circlular cross-sectional area. By differentiating this answer, we finally obtained an explicit formula for the surface area of the sphere. Please read the notes, and when you're ready, please attempt the exercises. Thank you very much for watching, and I look forward to seeing you again soon. [MUSIC]