Area of Polar Curve Region: Finding the Error

[ILoveBaseball]

Find the area of the region which is inside the polar curve r =5*cos(\theta)


and outside the curve r = 3-1*cos(\theta)

when i plugged those two functions into my calculator and found the bounds from 1.0471976 to 5.2359878.

my integral:
\int_{1.0471976}^{5.2359878} 1/2*(5*cos(\theta))^2 - 1/2*(3-1*cos)^2
and got a final answer of
-4.10911954 which is incorrect. anyone know what i did wrong?

 

[AKG]

You can't simply plug stuff into your calculator, then plug numbers into a formula without thinking and expect to get the right answer. The first thing you should always do is draw yourself a picture. The first fuction looks like a teardrop, with the point on the origin, and the axis of the tear drop being the x-axis, and the drop getting fatter to the right, and then rounding off. Key points (in Cartesian co-ordinates) will be (0,0) and (5,0). Also, this function will repeat its behaviour after pi. It will trace out a tear drop when theta goes from 0 to pi, and then trace over that same path again from pi to 2pi.

The second graph looks like an egg. You can think of an egg having a fat bottom and narrowing at the top. For this egg, the top will be to the left of the y-axis, and the bottom will be to the right (it's an egg lying on its side, with axis being the x-axis again). Key points: (2, 0), (0, 3), (-4, 0), (0, -3).

You should be able to see two points of intersection, one in quadrant I (x> 0, y>0), the other in quadrant IV (x>0, y<0). You know that intersection only happens when both curves have the same r values AND the same theta values, or the r value of one is the negative of the other AND the theta value of one is that plus pi of the other. You will only satisfy this condition when cos(theta) = 1/2. You should know that theta = 60 degrees here. Or you might remember that sin(30 degrees) = 1/2 and should be able to tell from this that cos(60 degrees) will be 1/2, i.e. you should never use a calculator (do they even allow to do so for tests anyways?) You can easily change 60 degrees into radians. Also, if cos(theta) is positive (like 1/2), then you can find values in quadrants I and IV for theta, so your other point of intersection will occur at -60 degrees.

Actually, you don't need that other point of intersection. You should see by symmetry that you can find the area by finding 2 x another area, where theta goes from 0 to pi/2. From 0 to pi/3, you will find the area under the second curve, and from pi/3 to pi/2, you will find the area under the first curve. Sum these, multiply by 2 and get your answer.

You can and should do this all without a calculator, and I can't imagine how you could do this without drawing pictures and thinking about it and just using a calculator. You should draw a picture, and verify the statements I've made:

1) The first fuction looks like a teardrop, with the point on the origin, and the axis of the tear drop being the x-axis, and the drop getting fatter to the right, and then rounding off. Key points (in Cartesian co-ordinates) will be (0,0) and (5,0).
2) The second graph looks like an egg. You can think of an egg having a fat bottom and narrowing at the top. For this egg, the top will be to the left of the y-axis, and the bottom will be to the right (it's an egg lying on its side, with axis being the x-axis again). Key points: (2, 0), (0, 3), (-4, 0), (0, -3).
3) You should be able to see two points of intersection, one in quadrant I (x> 0, y>0), the other in quadrant IV (x>0, y<0).
4) You know that intersection only happens when both curves have the same r values AND the same theta values, or the r value of one is the negative of the other AND the theta value of one is that plus pi of the other.
5) You will only satisfy this condition when cos(theta) = 1/2. Note you are satisfying two conditions here with cos(theta), the first one where r and theta values are the same, the second where they are opposite (and you should be able to see why theta is the opposite of pi + theta)
6) You should see by symmetry that you can find the area by finding 2 x another area, where theta goes from 0 to pi/2
7) From 0 to pi/3, you will find the area under the second curve, and from pi/3 to pi/2, you will find the area under the first curve.

 

[thepatientmental]

It is difficult to determine exactly what went wrong without seeing the steps you took in your calculation. However, there are a few potential sources of error that could have led to an incorrect answer.

Firstly, it is important to make sure that the bounds of the integral are correct. In this case, the bounds should be from 0 to 2π (the full range of θ values for the given curves). It appears that you used the bounds 1.0471976 and 5.2359878, which are about 60 and 300 degrees, respectively. This may have led to an incorrect answer as the integral would not have covered the entire area.

Another potential source of error could be in the setup of the integral itself. It is important to carefully consider the limits and the functions being integrated. In this case, the functions being integrated should be the area between the two curves, which can be represented as (1/2)r^2. Therefore, the integral should be set up as follows:

\int_{0}^{2\pi} 1/2*(5*cos(\theta))^2 - 1/2*(3-1*cos(\theta))^2 d\theta

Additionally, make sure to use the correct form of the cosine function in your calculator. Some calculators may require the use of parentheses around the argument, so it should be entered as 5*cos(\theta) and 3-1*cos(\theta).

Overall, it is important to double check your bounds and setup of the integral, as well as the use of the correct form of the function in your calculator. If the issue persists, it may be helpful to consult with a tutor or instructor for further clarification.