How does the placement of a point affect solving a polar integral?

[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 shouldn't i solve it the normal way as if the point wasnt mentioned at all.

 

[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.

 

[transgalactic]

what shaded area
i see only a formula no shades

about the second question
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??

 

[transgalactic]

i know how polar works but..

but i still don't see in what way i am suppose to know the rangle
only by looking at the formula

 

[unplebeian]

but i still don't see in what way i am suppose to know the rangle
only by looking at the formula

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.

 

[HallsofIvy]

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, [itex]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", $\theta+ \pi$ and that can confuse things. When $\theta= 0$, r= 2a. When [itex]\theta= $\pi/2$, 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.

 

[transgalactic]

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

and that we need to find the are of the field 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??

 

[HallsofIvy]

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

and that we need to find the are of the field 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".

 

[unplebeian]

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

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