Okay, let us consider this example of a half circle with radius R. The coded y clearly is a function of x since x square plus y square is equal to r square. The functional expression for y of x is square root of R square minus x square. Now, let's say that we plan to use Cartesian coordinates and we plan to use a differential element dxdy, then the area of this circle can be expressed as a double integral of dxdy, where x goes from minus R to R, and y goes from zero to square root of R square minus x square. We integrate over y, so we get y, so x from minus R to R of square root of R square minus x square. Integral of the area of dy we said is y, so we get this times dx. However, this integral cannot be calculated easily analytically. So also in case we had chosen to use a vertical strip, what would have happened? We simply would have skipped the first integration and we would have gone directly to this point because indeed in this case, the area of this little strip dA is the width which is dx and the height which is square root of R square minus x square. So Cartesian coordinates, or vertical, or horizontal strips do not work in this case. So in general, whenever we have boundaries which are described by circle or ellipse, it is preferable to use polar coordinates. Let's use polar coordinates. Let's assume we have polar coordinate r and theta, we use a differential element as described earlier where this side is dr this is r the theta, and we write this as a double integral. Theta goes from zero to Pi, and r goes from zero to R of this differential element which is a little rectangle with two sides dr and d theta approximately, so it is r, dr, d theta. Integral of rdr gives us square. So what we have is integral from theta zero to pi, R square over 2 from zero to r gives R square over 2 times d theta. This gives R square over 2 times Pi. So pi R square over 2, the area of this half circle something we knew. By the way, again when we started with a little infinitesimal element in polar coordinates, we could have skipped this first step and use directly the larger infinitesimal element with area dA corresponding to this larger widths with an angle d theta and the area of this width is approximately the area of a triangle with the base of this is R d theta, the height is R so the area is 1.5 R square d theta. So this is the area of this larger triangle we see here. It is possible to have used a different infinitesimal quandary of first order. For example, we could have used a little course strip like this, where the thickness of the strip is dr and it is a half circle. This strip if I call this area dA, then I could write this a one-dimensional integral, my R goes from zero to R, and the area dA of this strip is the length of the strip times this thickness. The length of the strip is Pi r, the thickness of the strip is dr. So we get Pi 1.5 r square from zero to R, so we get again pi R square over 2 which is the correct area. Now, let us go and try to calculate now the centroidal coordinates. That is, if there's a centroid, the first thing we observe is that in this case the centroid c must be lying along the y-axis. This is always true whenever you have a shape which has an axis of symmetry. Whenever we have an axis of symmetry, for example like this, then the centroid must lie on the axis of symmetry. There is an an example, that in this case for every area dA, there is a symmetric area dA on the other side such that this point has a coordinate x and this has a coordinate minus x. So when we integrate the integral of x times dA in this case is zero, because these two cancel out. So this exactly case here, we have y is an axis of symmetry and therefore x is equal to zero. So we will only try to calculate the yc in this case. So to calculate the centroidal distance yc in this case, we will not use Cartesian coordinates because we already said in the case of Cartesian coordinates, the boundaries create problems and therefore we will use as before the polar coordinates. So we consider this element. This element has the height of this element here. The distance y is equal to r Sine of theta, and therefore the first moment of the area. That means the first moment y times dA in this case becomes r Sine of theta times the little area dA which in this case is r, dr, d theta. We have a double integration where r goes from zero to R and theta goes from zero to Pi. Now, when we integrate over r, we get R squared dr gives R cubed over 3, so we get from zero to R, so we get R cubed over 3 times Sine of theta d theta. Now, this is a one-dimensional integral over theta and this is what we would have gotten if we would have considered directly and infinitesimal area of first order that is a little weights with an angle d theta. In this case, this area is 1.5, the height is R, the base is r d theta so 1.5 R squared d theta, same as we talked about earlier here and the distance y we should use if we consider this larger area dA, if we consider this is dA, then the y could be anything from zero to R but it must be the number that represents the centroid of this little area. We know that the centroid of a triangle lies at two-thirds based on what we said earlier for triangle, we proved that the centroid is at two-thirds of the distance, so two-thirds of this distance. So therefore, if we were going to use from the beginning the larger infinitesimal element, we would have written the centroid of this here, so that distance y that we could use in this case would be the height of this point, that is two-thirds of R times Sine of theta, times the area which is 1.5 R square d theta. Now, two and two cancels out, this gives R cubed so we get exactly the same expression, r cubed over 3 Sine of theta d theta. Now, this integral, we can continue to work it out now. Integral of Sine theta d theta is minus Cosine of theta so we have r cubed over 3 minus Cosine of theta from zero to Pi, which gives two-thirds of R cube. Therefore, the centroidal distance yc is the first moment over the area. So two-thirds of R cube divided by the area which was Pi R square over 2, and this gives 4 over 3 pi times R. So this distance is 4 over 3 Pi, that's approximately 0.424 R. So what we showed is that the centroid of half a circle lies at a distance of less than half of R. That means the centroid is closer located to the center then to this end. That makes sense because the lower part of this body has bigger contributing areas than the upper part which is smaller areas. So the centroid is shifted towards to represent the fact that it is a weighted average of the distance of the areas and therefore it turns out to be this number which is less than 0.5 R in this case. So we showed that in the case of shapes which have circular boundaries, we should use polar coordinates. It's up to us if we want to use information quantities of second or first order, if we use the first order quantities, we skip one of the integrations but we have to be careful when we use in the first moment the expression for y to use the y if y is not uniquely defined for the area we consider we have to use the y corresponding to the centroid of the area we are considering.