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arildno

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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}+\bigtriangleup{\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?

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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)

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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

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arildno

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You do know how to integrate, don't you?n05tr4d4177u5 said: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)

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.

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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

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arildno

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y = d(theta) Y ^ 2

integrate from y=R to y= 0

Shape is semi circle

integrate from y=R to y= 0

Shape is semi circle

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arildno

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

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R = radius, integrate from 0 to R

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arildno

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What is Y?

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

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

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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)

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arildno

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n05tr4d4177u5 said:in a rectangles case, dA = dX dY

if i have the rectangles width and it is 4 m, then dA = 4dY,

Well, why do you believe this to be true?

How do you derive that?

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

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example , the integral of ydy = y squared/ 2

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arildno

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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

Integral

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You are good up to here.n05tr4d4177u5 said: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,

Why can you write this? Where did it come from? You cannot just drop the dx and it is not automaticlly set to 4 or any other value.

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]

Can you complete the above integral?which lets me integrate along y axis,

now for a semicircle, how would i do that if dA = r dr d(theta)

Ok, now lets set up the integral for the area of a semi circle.

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?

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HallsofIvy

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R = radius, integrate from 0 to R

You haven't stated a problem here!

If you just want the area, look at the problem you gave:

(And arildno did not say he had never seen that before, he was trying to get you to think about why you think that was true.)in a rectangles case, dA = dX dY

if i have the rectangles width and it is 4 m, then dA = 4dY

Suppose your semi-circles 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

[tex]\int_{\theta=0}^{\pi} R^2 d\theta= \pi R^2[/tex], exactly the right answer.

(θ does

(Oh, and by the way, why was this posted under "differential equations". Was this part of a differential equations problem?)

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