Finding the Area Enclosed by a Polar Curve

In summary, the goal of this problem is to find the area enclosed by the polar curve r = 2e^(0.9theta) on the interval 0 <= theta <= 1/8 and the straight line segment between its ends. The boundaries of integration are given as a=0 and b=1/8, and the formula for finding area in polar coordinates is used to set up the problem. The correct integral to solve is (1/2*(2 e^(0.9theta))^2)dTheta.
  • #1
HolyDesperado
16
0

Homework Statement



Find the area enclosed by the polar curve
r = 2 e^(0.9theta)

on the interval 0 <= theta <= 1/8
and the straight line segment between its ends.

Homework Equations



arclength =
eq0006MP.gif


The Attempt at a Solution



I need help finding the boundaries for this problem. Is it from 0 to pi? I'm not sure how exactly would I go about finding the limits of integration(boundaries) since it starts from the origin and ends at the x-axis.

http://img5.imageshack.us/img5/7910/whatareloi.jpg [Broken]
 

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  • #2
The integration boundaries for theta are given.

[Hint: first find the area between the curve and the x-axis, then find the area under the straight line segment between its ends and subtract appropriately.]
 
  • #3
CompuChip said:
The integration boundaries for theta are given.

[Hint: first find the area between the curve and the x-axis, then find the area under the straight line segment between its ends and subtract appropriately.]

are you talking about the inequality from 0 to 1/8? so my boundaries of int. are a=0, b=1/8?
 
  • #4
Yes.
Note that the dtheta is outside the square root and don't forget the Jacobian factor when doing an integral in polar coordinates.
 
  • #5
CompuChip said:
Yes.
Note that the dtheta is outside the square root and don't forget the Jacobian factor when doing an integral in polar coordinates.

integrating from 0 to 1/8 is giving me .355, which isn't right =-\
 
  • #6
Why did you include an arclength formula in your OP when the question asks for area?
 
  • #7
i thought it meant arclength, but i guess it is area
so it should be int [a->b] of (1/2(r)^2)dthetabut I still don't know how to set up this problem correctly
 
  • #8
I figured it out, it's Integral[0->1/8] (1/2*(2 e^(0.9theta))^2)dTheta

thanks everyone
 

1. What is a polar curve?

A polar curve is a type of curve that is defined by a mathematical equation using polar coordinates. Unlike a traditional Cartesian coordinate system, polar coordinates use a distance from the origin and an angle to describe a point on a plane.

2. How do you find the area enclosed by a polar curve?

To find the area enclosed by a polar curve, you first need to graph the curve and determine the bounds of the area. Then, you can use the formula A = 1/2 ∫(r^2)dθ, where r is the radius of the curve and θ is the angle of rotation.

3. Can you use the same method to find the area for any polar curve?

Yes, the same method can be used to find the area for any polar curve. However, the process of determining the bounds and integrating the formula may vary depending on the complexity of the curve.

4. Is there a shortcut for finding the area enclosed by a polar curve?

Yes, there is a shortcut for finding the area enclosed by a polar curve called the Polar Area Formula. This formula is A = 1/2 ∫(r^2)dθ, where r is the radius of the curve and θ is the angle of rotation. This formula can be used for any polar curve without having to determine the bounds.

5. What is the significance of finding the area enclosed by a polar curve?

Finding the area enclosed by a polar curve can be used to solve real-world problems involving circular or spiral shapes. It is also a fundamental concept in calculus and can be applied to more complex mathematical problems.

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