Set up the polar integral that represents the area of the region R in the xy-plane enclosed by the circle x^2 + y^2 = 4, above the line y = 1, and below the line y = √3x.
Understand the Problem
The question is asking us to set up a polar integral that represents the area of a specific region in the xy-plane, defined by a circle and two lines. Our approach will involve identifying the limits of integration and the appropriate polar coordinate transformations for the given shapes.
Answer
$$ A = \int_{0}^{\frac{\pi}{3}} \int_{\frac{1}{\sin(\theta)}}^{2} r \, dr \, d\theta $$
Answer for screen readers
The polar integral representing the area is:
$$ A = \int_{0}^{\frac{\pi}{3}} \int_{\frac{1}{\sin(\theta)}}^{2} r , dr , d\theta $$
Steps to Solve
- Convert the circle equation to polar coordinates
The equation of the circle is given by $x^2 + y^2 = 4$. In polar coordinates, this becomes:
$$ r^2 = 4 \implies r = 2 $$
This represents a circle of radius 2.
- Identify the line equations in polar coordinates
The lines are defined as $y = 1$ and $y = \sqrt{3}x$.
- For the line $y = 1$:
In polar coordinates, this is:
$$ r \sin(\theta) = 1 \implies r = \frac{1}{\sin(\theta)} $$
- For the line $y = \sqrt{3} x$:
In polar coordinates, we have:
$$ r \sin(\theta) = \sqrt{3} r \cos(\theta) $$
which simplifies to:
$$ \tan(\theta) = \sqrt{3} \implies \theta = \frac{\pi}{3} $$
- Set up the limits of integration
Now we need to determine the bounds for $r$ and $\theta$.
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The angle $\theta$ will range from $\theta = 0$ (along the positive x-axis) to $\theta = \frac{\pi}{3}$ (the line $y = \sqrt{3}x$).
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The radius $r$ will range from the line $y = 1$ to the circle. The lower limit for $r$ is given by $\frac{1}{\sin(\theta)}$, and the upper limit is the circle defined by $r = 2$.
- Write the integral
The area $A$ can be expressed with the polar area integral:
$$ A = \int_{\theta=0}^{\frac{\pi}{3}} \int_{r=\frac{1}{\sin(\theta)}}^{2} r , dr , d\theta $$
- Evaluate the inner integral
The inner integral is:
$$ \int r , dr = \frac{1}{2} r^2 $$
Evaluating from $r = \frac{1}{\sin(\theta)}$ to $r = 2$, we can substitute these limits to find the area.
The polar integral representing the area is:
$$ A = \int_{0}^{\frac{\pi}{3}} \int_{\frac{1}{\sin(\theta)}}^{2} r , dr , d\theta $$
More Information
This setup captures the region outlined by the constraints of the circle and the lines in the xy-plane. The radius extends from the line $y=1$, represented in polar coordinates, up to the boundary of the circle $r=2$. The angular part covers the area from the positive x-axis to the specific line defined by $y = \sqrt{3}x$.
Tips
- Confusing the Limits of Integration: Be careful when setting the limits for both $r$ and $\theta$. Ensure that the bounds represent the correct geometrical features.
- Incorrectly Converting Equations: Always double-check the conversion of Cartesian coordinates to polar coordinates, especially with sine and cosine functions.
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