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Semicircle problemby n05tr4d4177u5
Tags: semicircle 
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#1
Sep1805, 09:20 AM

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i have been told that dA=R dR d(theta), what does mean exactly in terms of pi. the shape is a semicircle. Does this mean that d(theta) is 180, or pi. Please give me an example.



#2
Sep1805, 09:30 AM

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No, dA is a notation for an area element (infinitely small).
1. Suppose you are using polar coordinates [tex]r,\theta[/tex], what sort of region is described by the inequalities: [tex]R\leq{r}\leq{R}+\bigtriangleup{R},\theta_{0}\leq\theta\leq\theta_{0}+\b igtriangleup{\theta}[/tex] You ought to see that this represent a specific portion of a section of a circle. 2. Now, if you let [tex]\bigtriangleup{R},\bigtriangleup\theta[/tex] tend towards zero, you may approximate this region as a rectangle (or trapezoid, if you like that better). What should the (tiny!) area of this figure be? 


#3
Sep1805, 09:53 AM

P: 14

here is an example of mine that i have
suppose i hae a rectangle, in a rectangles case, dA = dX dY if i have the rectangles width and it is 4 m, then dA = 4dY, which lets me integrate along y axis, now for a semicircle, how would i do that if dA = r dr d(theta) 


#4
Sep1805, 09:55 AM

P: 14

Semicircle problem
so what i did, was changed dr to dy, and now i need to change d(theta) which should be a number (like in the case of 4 m for rectangle), and it has to be in terms of pi,
the shape is a semi circle 


#5
Sep1805, 09:57 AM

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If you are to compute the area of the semicircle, the [tex]\theta[/tex] integration is trivial, so you can find out that by yourself. 


#6
Sep1805, 10:02 AM

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no i dont know, its been a while since i integrated a circle
please help me, thats why im asking, i havent done that in years 


#7
Sep1805, 10:06 AM

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Well, can you at least post the integral you want to compute, specifying the intervals in which R and the angle must lie in for the semicircle?



#8
Sep1805, 10:09 AM

P: 14

y = d(theta) Y ^ 2
integrate from y=R to y= 0 Shape is semi circle 


#9
Sep1805, 10:11 AM

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This is not what is in your book (it doesn't make any sense). Please type in correctly.



#10
Sep1805, 10:20 AM

P: 14

imagine the top half of a circle. the origin lies along the bottom of the semicircle, and in the middle. y axis up, and x axis to the right and left. i think theta can only go from 0 to 180 degrees since it is a semi circle. Y = d(theta) R squared
R = radius, integrate from 0 to R 


#11
Sep1805, 10:27 AM

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What is Y?
Do you know what the d in front of (theta) means? 


#12
Sep1805, 10:28 AM

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in a rectangles case, dA = dX dY
if i have the rectangles width and it is 4 m, then dA = 4dY, which lets me integrate along y axis, now for a semicircle, how would i do that if dA = r dr d(theta) 


#13
Sep1805, 10:31 AM

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How do you derive that? 


#14
Sep1805, 10:35 AM

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are u reallly an advisor, cuase its unbelievable youve never seen that before,



#15
Sep1805, 10:37 AM

P: 14

example , the integral of ydy = y squared/ 2



#16
Sep1805, 10:38 AM

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Sure I have, but you are so vague and obtuse that I don't think you understand what you write yourself.
So, again: show that you have at least a minimum of mathematical skill and understanding. I have already shown you how you can do this problem, but it doesn't seem you have the competence to follow a perfectly clear line of reasoning. 


#17
Sep1805, 03:52 PM

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One of the first things you need to do is carefully define your rectangle in terms of your coordinate system. Suppose one corner is at the origin, x=y=0 and the diagonal corner is at x=L, y=W. To find the area we must define our integral: [tex] \int dA = \int^{y=W}_{y=0} \int^{x=L}_{x=0} dx dy [/tex] Once again we need to define the circle in terms of the coordinate system. If the center of the circle is at the origin and the radius is R. So to get the Area of the circle (or portion there of) the radius will take on values between 0 and R, while the angle [itex]\Theta[/itex] varies from 0 to [itex] \pi[/itex] (for a semi circle). Now lets set up the integral [tex] \int dA = \int ^{\Theta=\pi} _{\Theta=0} \int^{r=R}_{r=0} rdr d\Theta[/tex] Can you evaluate the integral? 


#18
Sep1805, 06:51 PM

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If you just want the area, look at the problem you gave: Suppose your semicircles radius is R. Then dA= r dr dθ. In a sense, dr here, as well as r, is r: along each radius we go from r= 0 to R so the "change in r", dr, is R. Okay, then r dr dθ becomes dA= R^{2} dθ, just like your dX dY became 4dy, the area of the semicircle is [tex]\int_{\theta=0}^{\pi} R^2 d\theta= \pi R^2[/tex], exactly the right answer. (θ does not go from 0 to 180! As I am sure you learned in when you were learning the derivatives of sine and cosine, in calculus, all angles are in radians.) (Oh, and by the way, why was this posted under "differential equations". Was this part of a differential equations problem?) 


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