# Polar integral questions

1. Jan 12, 2008

### transgalactic

the first one is

find the area around this formulas

1. r=a(1+cos(teta))
i know the formula
my problem is,
how am i suppose to know what is the range of the angles of the integral
??

2.
r=a(cos(teta)+sin(teta))
where there is a point (a/2,0)
inside the area.

how is that point sopposed to change the way i am solving it
why shouldnt i solve it the normal way as if the point wasnt mentioned at all.

2. Jan 12, 2008

### unplebeian

1. You need to look at how the plot is going to look like. For values of theta= 0..2Pi plot r. e.g. If r= 1, (here there is no theta) but the plot is a circle of rad=1.

Polar plots are difficuit to plot so I would suggest polarplot function in Maple

2. After looking at the plot, I see that point (a/2,0) is point (a/2,0) on the x-y axis.

See attached.

#### Attached Files:

• ###### polarplots.JPG
File size:
9.5 KB
Views:
70
Last edited: Jan 12, 2008
3. Jan 12, 2008

### transgalactic

i see only a formula no shades

i was given a formula
r=a(cos(teta)+sin(teta))
generally i whouls solve it with a given formula
(if i knew what angles to put in the integral)

but they add another info that puzzles me
"where there is a point (a/2,0)
inside the area"

what does it meen??

4. Jan 13, 2008

### transgalactic

i know how polar works but..

but i still dont see in what way i am suppose to know the rangle
only by looking at the formula

5. Jan 13, 2008

### unplebeian

You're right. You cannot have any clue of the range by just looking at the formula. You have to plot it. That's the only way. Plotting polar graphs is hard, so I suggest use of a software such as Matlab or Maple which will give you some insight initially.

6. Jan 13, 2008

### HallsofIvy

Staff Emeritus
You don't really need to do a very accurate graph. Since sine and cosine are periodic with period $2\pi$ You can "check" submultiples of that.

The first one, $r= a(1+ cos(\theta)), is particularly easy since r is never negative (assuming a> 0). Since a distance is always positive, the graph when r< 0 is interpreted as "going the other way", [itex]\theta+ \pi$ and that can confuse things. When $\theta= 0$, r= 2a. When $\theta= [itex]\pi/2$, r= a, when $\theta= \pi$, r= 0, when $\theta= 3\pi/2$, r= a, and when $\theta= 2\pi$, r= 2a again. After that, it traces the graph over again. In order to go around the graph a single time, $\theta$ can go from 0 to $2\pi$.

For the second one, $r=a(cos(\theta)+sin(\theta))$, you need to be more careful.

7. Jan 13, 2008

### transgalactic

about the second one i was tald that there is two common feilds

and that we need to find the are of the feild for which the given point exists

so i will try to buil them by entering point of 0 pi/2 pi p*1.5 2pi
is this the universal way to plot a graph??

8. Jan 14, 2008

### HallsofIvy

Staff Emeritus
and that we need to find the are of the feild for which the given point exists[/quote]
Did you write the equation correctly then?

If $r= a(cos(\theta)+ sin(\theta))= acos(\theta)+ asin(\theta)$, then, multiplying both sides by r, $r^2= arcos(\theta)+ arsin(\theta)$ or
$x^2+ y^2= ax+ ay$ so $x^2- ax+ y^2- y^2= 0$

Completing the square, $x^2- ax+ a^2/4+ y^2- ay+ a^2/4= a^2/2$ so $(x-a/2)^2+ (y-a/2)^2= a^2/2$.

The graph is a circle with center at (a/2, a/2) and radius $a/\sqrt{2}$, not two different "fields".

9. Jan 14, 2008

### unplebeian

(a/2,0) in (r,theta) is (a,a/2) on the x-y axis