My Math Forum Polar function from a (circle and square) shape

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 June 13th, 2013, 04:27 PM #1 Member   Joined: Dec 2010 Posts: 51 Thanks: 0 Polar function from a (circle and square) shape http://i.imgur.com/sUM8fwH.gif I'm trying to find the area of the blue shape (technically i'm trying to find the centroid). How would i setup the integral in polar form? $\int_{\pi}^{3\pi/2} \ \ \ \int_{?}^{?} drd\theta$
 June 13th, 2013, 06:50 PM #2 Senior Member   Joined: Jan 2013 Posts: 113 Thanks: 0 Re: Polar function from a (circle and square) shape If the image is accurate, so that the the axes are at a tangent to the circle, then an integral is not really necessary. We have a quarter circle and a square. $\frac{1}{4}A_{circle}= \frac{1}{4}\pi R^2$ $A_{square}= R^2$ Hence the area shown is $A_{blue}= A_{square}-\frac{1}{4}A_{circle} = R^2(1-\frac{\pi}{4})$ If you want to do it with integration, just do the same thing. Consider the graph illustrating the two functions: $r(\theta)=R$ $y(x)=-R$ Then, being explicit with your request, we can find the area of the bottom-left quadrant of the circle described by $r(\theta)$: $\frac{1}{2}\int^{3\pi/2}_{\pi}r^2 d\theta= \frac{R^2}{2}\int^{3\pi/2}_{\pi}1 d\theta = \frac{R^2}{2}\left[\theta\right]^{3\pi/2}_{\pi} = \frac{1}{4}\pi R^2$ And the square is just the area bound between the line y(x) and the x-axis from -R to 0. Notice that since the line is below the axis, we're going to get the negative area. To correct, we'll just multiply by -1: $-\int_{-R}^0 -R dx= R\left[x\right]^0_{-R} = R^2$ And then the area of the blue is the square minus the quarter circle: $A_{blue}= A_{square}-\frac{1}{4}A_{circle} = R^2(1-\frac{\pi}{4})$ To find the centroid of an object with upper bound f(x) and lower bound g(x) over the region $a\leq x\leq b$, the following formulae can be applied to find the centroid $(\bar{x},\bar{y})$: $\bar{x}=\frac{1}{A}\int_a^b x[f(x) - g(x)]\;dx$ $\bar{y}=\frac{1}{A}\int_a^b \left[\frac{f(x) + g(x)}{2}\right][f(x) - g(x)]\;dx$ Where A is the area of the shape enclosed by f and g. Formulae thanks to wiki. If we let: $f(x)=\sqrt{R^2-x^2}$ And $g(x)=-\sqrt{R^2-x^2}$ Then it's clear the area enclosed is $\pi R^2$, since f and g are two halves of a circle with radius R. We will look at the region -R
 June 14th, 2013, 02:00 PM #3 Member   Joined: Dec 2010 Posts: 51 Thanks: 0 Re: Polar function from a (circle and square) shape thank you very much

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### how to find centroid of circle &half circle

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