 There's another way that we can use to look at the volume of a cylinder First of all we we are using double integrals, and we're just using the xy coordinate Cartesian coordinate system We can convert that to a polar grid If we do the conversion to the polar grid it might be easier for us to do this remember we were still using the fact that z equals the f of some vector function or of x and y equals some constant h So we have this cylinder going up to a flat plane and we want to denote We want to get to an equation for this of volume using just double integrals now instead of using then the fact that we have a type 1 region going from this y equals 0 up until this y equals the square root of a squared minus x squared Where we have the fact that this outer cylinder the outer wall of the cylinders x squared plus y squared equals some radius called a squared So this whole radius here would be till a Now the same as we did with normal Let's do it this way remember we are doing our double integral over some region at the bottom Let's call it region R our function is just a constant over some area and Because of the type 1 region we call da equals dy dx That is how we made these infant tests to some really small little Blocks and we sum over all of them that is an integral But now we're going to construct a different tiny little rectangle block and here I have to note to them. We've scaled it up in as much as we have this tiny little angle d theta and This tiny little angle d theta can denote if it's very small it approximates the width of this little rectangle of us of ours and If we look on this going from zero till the radius going from zero till a we have a here If we take just an infinitely small part of that and we call that DR If we call that DR, that'll be the height of our little rectangle Now we just need the width of our rectangle and what is the equation for and for arc length? remember arc length Equals just the radius times The little angle that you the angle that you meeting theta 2 minus theta 1 equals delta theta That is certainly what we are going for So that is arc length and if it becomes infinitesimally small this approximates a straight line So we still have height times width So what would be this area? The area now of this tiny little block becomes height times width That is r d theta Not delta theta for the difference between the two becomes infinitesimally small remember It's it's d theta it's d dr. So it's r d r d theta R d theta those two and d r the length and the width of that little rectangle So now we have over some region of This constant function of ours are d r d theta you can also do d theta d r doesn't matter Now we just have to look at the bounds of these two of This new grid of ours So the first thing to convert to a new grid is to start with the basics integration is forming infinitesimally little small line segments or Little rectangles then it's called a little rectangles. That's what we did with the Riemann sums when we started learning about integration We're still constructing these tiny little areas. It has a width It is hey a height if we multiply those two by each other Which is the equation for a rectangle height times width width times length, whichever way you look at it And just to remember that this is r times d theta And this is dr multiply them and you get area. So For Cartesian coordinate system, it was this x times y dy dx So now we have our dr d theta. So let's just work out what the bounds are going to be we'll take conventionally. We'll put the R In the middle because it usually is a constant now. What happens to the radius it goes from a radius of zero to the radius of a So that becomes easy the inner integral here The inner integral becomes very easy In as much as as far as h. There's a r dr the dr becomes From zero till a and on the outside. We just have to think what happens to this angle theta Well, we're going from theta equals zero all the way around to two pi That's what's happening. We're sweeping this tiny little blocks all the way around and all the way from the center out and if we add then all these Uncountable little rectangles we are integrating so that goes from zero to two pi and that is going to be the volume of this cylinder of ours between this plane at the bottom and a y-axis which we haven't drawn here Equaling z equals h or the f of x and y equals h And there we have it with a d theta So this is going to equal zero to two pi. We can take the h out Going from zero to a of r dr, and then the outside function d theta From zero to two pi of h. What is this? It's going to be a half r squared going from zero to a d theta We can certainly bring this h out now also. We can bring the half out as well and We can this is going to be a squared minus zero squared Now the a is also a constant so that can also come out and all we left with is the zero to two pi d theta That's all we have left. That's going to be an h over two That is going to be an a squared and that's gonna be a theta going from zero to two pi in other words an H over two and a squared and a two pi Eventually, what are we left with? We are left with Pi a squared to h and remember this a represented the radius that is pi r squared h And that is the equation of the volume of a cylinder Changing from a Cartesian coordinate system to a polar coordinate system Just remembering the only difference we did is to calculate at the area of tiny little infinitesimally little tiny little rectangles in a different way