What is the integral setup for finding probability within a disk?

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To find the probability of a point uniformly distributed within a disk of radius 1, the integral setup involves integrating over a disk of radius x. The correct density function is f(x,y) = C for 0 ≤ x² + y² ≤ 1. The discussion highlights the need for proper limits of integration, which can be effectively addressed using polar coordinates. In polar form, the integral simplifies the setup and clarifies the relationship between the variables involved. Understanding these mechanics is crucial for correctly setting up the probability calculation.
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I'm taking a probability class where multivariate calculus was not a prerequisite, but some of it is coming up, I get the concept of, say integrating over a region, but get lost in some of the mechanics

Here is the problem (I don't want a full solution):


A point is uniformly distributed within the disk of radius 1. That is its density is
<br /> f(x,y) = C \hspace{1cm} 0 \leq x^2 + y^2 \leq 1<br />
Find the probability that its distance from the origin is less than x, 0 \leq x \leq 1

I'm pretty sure I have to set up an integral that integrates over a disc of radius x to get the probability
Something like this
<br /> \int_A \int_B C \, dx \, dy<br />
But I don't know what the intervals A and B are supposed to be.

Can someone point me in the right direction? I get confused because my attempts end up with x being involved in the limit of integration, but x is the dummy variable, which doesn't seem right.
 
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Bre Ntt said:
I'm taking a probability class where multivariate calculus was not a prerequisite, but some of it is coming up, I get the concept of, say integrating over a region, but get lost in some of the mechanics

Here is the problem (I don't want a full solution):


A point is uniformly distributed within the disk of radius 1. That is its density is
<br /> f(x,y) = C \hspace{1cm} 0 \leq x^2 + 1 \leq 1<br />
Shouldn't the inequality be 0 <= x2 + y2 <= 1?
Bre Ntt said:
Find the probability that its distance from the origin is less than x, 0 \leq x \leq 1
Bre Ntt said:
I'm pretty sure I have to set up an integral that integrates over a disc of radius x to get the probability
Something like this
<br /> \int_A \int_B C \, dx \, dy<br />
But I don't know what the intervals A and B are supposed to be.

Can someone point me in the right direction? I get confused because my attempts end up with x being involved in the limit of integration, but x is the dummy variable, which doesn't seem right.
 
Yes, sorry. 0 <= x^2 + y^2 <= 1

I fixed it above.
 
This would be a natural for polar form of a double integral.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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