Average Value and Area in Polar

[TranscendArcu]

Homework Statement

Let Q be the region bounded by r=sin(3θ) in the first quadrant. Find the area of Q. Find the average distance of points in Q from the origin.

The Attempt at a Solution

I thought I could calculate area like so:
\int_0 ^{pi/3} \int_0 ^{sin(3θ)} sin(3θ) dr dθ This gives answer π/6But if I calculate area via a single integral, I have,\int_0 ^{pi/3} (1/2) (sin(3θ))^2 dθ = π/12So clearly I am not doing something correctly in my calculation of the area.

I'm not quite sure how to begin finding the average distance, but I assume I'll have to divide by the area at the end.

 

[Mark44]

Homework Statement

Let Q be the region bounded by r=sin(3θ) in the first quadrant. Find the area of Q. Find the average distance of points in Q from the origin.

The Attempt at a Solution

I thought I could calculate area like so:
\int_0 ^{pi/3} \int_0 ^{sin(3θ)} sin(3θ) dr dθ

Your mistake is that you have sin(3θ) as both the integrand and as one of the limits of integration. Your integrand should be 1, and dA should be r dr dθ. With that change, after doing the inner integration, you get the same integral as below.

This gives answer π/6But if I calculate area via a single integral, I have,\int_0 ^{pi/3} (1/2) (sin(3θ))^2 dθ = π/12So clearly I am not doing something correctly in my calculation of the area.

I'm not quite sure how to begin finding the average distance, but I assume I'll have to divide by the area at the end.

 

[Simon Bridge]

What he said - also, I thought the first quadrant was 0-pi/2

 

[Mark44]

What he said - also, I thought the first quadrant was 0-pi/2

It is, but as θ ranges from 0 to \pi/3, 3θ ranges from 0 to \pi, hence r goes from 0 to 1 and then back to 0. Neither of the other two petals on this three-petaled rose is in the first quadrant, so the integration limits of 0 and \pi/3 are fine.

 

[vela]

I'm not quite sure how to begin finding the average distance, but I assume I'll have to divide by the area at the end.

r is the distance from the origin, so you want to calculate the average value of r. That's given by
\langle r \rangle = \frac{1}{A} \int r\,dAwhere A is the total area and dA is the area element Mark gave above.

 

[Simon Bridge]

It is, but as θ ranges from 0 to \pi/3, 3θ ranges from 0 to \pi, hence r goes from 0 to 1 and then back to 0. Neither of the other two petals on this three-petaled rose is in the first quadrant, so the integration limits of 0 and \pi/3 are fine.

Ah - serves me right for not following my own advice and sketching it :)