Calculating Area of Polar Curve: Help Needed!

In summary, the formula for calculating the area of a polar curve is ∫(1/2)r^2 dθ, and the bounds of integration can be found by setting the polar equation equal to 0 and solving for the values of θ at the points of intersection. A calculator can be used to find the area, but it is important to double-check the bounds and equation for accuracy. There is a difference between finding the area of a polar curve and finding the area under a polar curve, with the latter requiring a different formula and bounds. Special cases such as self-intersecting curves or curves with negative or undefined values may require adjusting the calculation in multiple parts.
  • #1
schaefer
1
0
I've got a test tomorrow and I can't figure this one out:

r[tex]^{2}[/tex] = 9cos(5[tex]\vartheta[/tex])

I picked the inteval from 0 to [tex]\pi[/tex]/2 and I keep getting 18/5 but the back of the book says the answer is 18. I think something is wrong with my interval. Help Please!
 
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  • #2
when the angle changed from 0 to pi/2 ,you DON'T get the whole curve!
Try drawing out the curve.Note, cos(5*angle) must be > 0 to give a real point on the curve.

when angle=0, r=3
when angle=18 degree, r=0hope this is helpful
 
  • #3


Hi there,

Calculating the area of a polar curve can be tricky, but don't worry, I am here to help! First of all, let's make sure we understand the formula correctly. The formula for finding the area of a polar curve is:

A = ½∫r^2 dθ

This means that we need to integrate the function r^2 with respect to θ, and then multiply by ½. So, let's start by setting up the integral for the given curve:

A = ½∫9cos(5θ) dθ

Now, we need to determine the limits of integration. The interval you chose, from 0 to π/2, is correct. However, we also need to consider the symmetry of the curve. Since we are only integrating from 0 to π/2, we are only considering the area in the first quadrant. But, since the curve is symmetric about the x-axis, we can multiply our result by 4 to get the total area. So, our final integral becomes:

A = 4 * ½∫9cos(5θ) dθ

= 2∫9cos(5θ) dθ

Now, let's solve the integral:

A = 2 * 9/5sin(5θ) from 0 to π/2

= 2 * 9/5 * (sin(5π/2) - sin(0))

= 2 * 9/5 * (1 - 0)

= 18/5

It looks like your integral is correct, but you forgot to multiply by 2 in the end. So, the final answer is indeed 18.

I hope this helps! Good luck on your test tomorrow. Remember to always consider the symmetry of the curve when determining the limits of integration for polar curves.
 

1. What is the formula for calculating the area of a polar curve?

The formula for calculating the area of a polar curve is ∫(1/2)r^2 dθ, where r is the polar equation and θ is the angle of rotation.

2. How do I find the bounds of integration for a polar curve?

The bounds of integration for a polar curve can be determined by setting the polar equation equal to 0 and solving for the values of θ that correspond to the points of intersection. These values will serve as the lower and upper bounds of integration.

3. Can I use a calculator to find the area of a polar curve?

Yes, many scientific calculators have a built-in function for calculating the area of a polar curve. However, it is important to double-check the bounds of integration and the equation being used to ensure accuracy.

4. What is the difference between finding the area of a polar curve and finding the area under a polar curve?

Finding the area of a polar curve involves calculating the area enclosed by the curve, while finding the area under a polar curve involves calculating the area between the curve and the origin. The latter often requires the use of a different formula, ∫r^2 dθ, and different bounds of integration.

5. Are there any special cases when calculating the area of a polar curve?

Yes, there are a few special cases to consider when calculating the area of a polar curve. These include curves that intersect themselves, curves that do not start and end at the same point, and curves with negative or undefined values. In these cases, the area may need to be calculated in multiple parts and adjusted accordingly.

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