Problem parameterising an off-center disk

[Amaelle]

Homework Statement

parametrisating an off center disk with a center o (0;3/2) and a raduis r=1/2

Relevant Equations

x=rcosθ
y-3/2= rsinθ

Greeting

(x-3/2)^2+y^2=1/4

x^2+y^2-3y+9/4=1/4
r^2-3*r*sinθ+2=0
Δ=9sinθ^2-1

r=[3rsinθ-sqrt(sinθ^2-1)]/2

is my approch correct?
thank you!

 

[Mark44]

Homework Statement:: parametrisating an off center disk with a center o (0;3/2) and a raduis r=1/2
Relevant Equations:: x=rcosθ
y-3/2= rsinθ

Greeting

(x-3/2)^2+y^2=1/4

This would be the Cartesian equation if the center were at (3/2, 0).

x^2+y^2-3y+9/4=1/4
r^2-3*r*sinθ+2=0

Perhaps you have a typo in the first equation of the circle. This equation looks OK.

Δ=9sinθ^2-1

No, your discriminant is wrong.

r=[3rsinθ-sqrt(sinθ^2-1)]/2

No. You are using the Quadratic Formula to solve for r. The expression on the right side should not include r.

is my approch correct?
thank you!

 

[Amaelle]

This would be the Cartesian equation if the center were at (3/2, 0).
Perhaps you have a typo in the first equation of the circle. This equation looks OK.
No, your discriminant is wrong.
No. You are using the Quadratic Formula to solve for r. The expression on the right side should not include r.

thank you

after correcting my mistakes we got
r1=[3sinθ-sqrt(sinθ^2-8)]/2 and r2=[3sinθ+sqrt(sinθ^2-8)]/2
but i go lost on how to choose the correct r because we have two constraints:
r must be positive and sinθ^2-8 also should be positive.

my other concern is how to choose the integral range for θ as we have an off center circle

 

[Mark44]

This was your incorrect discriminant from post #1:

Δ=9sinθ^2-1

after correcting my mistakes we got
r1=[3sinθ-sqrt(sinθ^2-8)]/2 and r2=[3sinθ+sqrt(sinθ^2-8)]/2

Your discriminant is still wrong, but in a different way. The quadratic equation you're working with is $r^2 - (3\sin(\theta))r + 2 = 0$. The coefficients are $a = 1, b = -3\sin(\theta), c = 2$.

r must be positive and sinθ^2-8 also should be positive.

Before deterimining this, you need to get the correct equations for r.

 

[Amaelle]

yes so sorry it was a typo but still

Δ=9sinθ^2-8

r1=[3sinθ-sqrt(9sinθ^2-8)]/2 and r2=[3sinθ+sqrt(9sinθ^2-8)]/2
I'm still in the same confusion

 

[Mark44]

yes so sorry it was a typo but still

Δ=9sinθ^2-8

r1=[3sinθ-sqrt(9sinθ^2-8)]/2 and r2=[3sinθ+sqrt(9sinθ^2-8)]/2
I'm still in the same confusion

Much better.
For the square root to produce a real number, we must have $9\sin^2(\theta) - 8 \ge 0$. This places restraints on the possible values of $\theta$. Also, the largest possible value of $9\sin^2(\theta)$ is 9, so the quantity in the radical is at most 1.

For $r_1, r_2$ to be nonnegative, work with the numerators of the two fractions, keeping in mind the previous restrictions on $\theta$.

my other concern is how to choose the integral range for θ as we have an off center circle

What is the integral you're working with?

 

[Amaelle]

i want to compute the surface of the disk using the integral, and because the circle is off center i know i can't use [0. 2pi] and because we have constraint on theta because of the discriminant I'm confused

 

[Mark44]

i want to compute the surface of the disk using the integral, and because the circle is off center i know i can't use [0. 2pi] and because we have constraint on theta because of the discriminant I'm confused

Do you mean the area of the disk? If so, then you already know that its area is $\frac \pi 4$.

Let's take things a step at a time. What constraints are there on $\theta$? You must have $9\sin^2(\theta) \ge 8$, right?

 

[Amaelle]

sin(x)>2√2/3 or sin(x) <-2√2/3 x>70.52° or x<-70.52°

 

[Mark44]

sin(x)>2√2/3 or sin(x) <-2√2/3 x>70.52° or x<-70.52°

Look at the graph of the sine function. In the first arch, i.e., between 0° and 180°, won't $70.53° < x < 109.47°$ satisfy $\sin(x) \ge \frac{2\sqrt 2}3$?

Regarding the two values of r produced by the Quadratic Formula, I suspect that the one with the + sign generates the upper half, more or less, of the disk, and the one with the - sign generates the lower part.

 

[pasmith]

i want to compute the surface of the disk using the integral, and because the circle is off center i know i can't use [0. 2pi] and because we have constraint on theta because of the discriminant I'm confused

The best parametrization of (x - x_0)^2 + (y - y_0)^2 \leq a^2 is <br /> \begin{split}x &amp;= x_0 + r\cos \theta,\\ y &amp;= y_0 + r\sin\theta \end{split} for 0 \leq r \leq a and 0 \leq \theta &lt; 2\pi.