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Homework:HW SECTION 12.3
Question 3, 12.3.41
Part 2 of 2
HW Score: 57.14%, 4 of 7 points
 Points: 0 of 1


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Part 1
a. Find all intersection points of the following curves.
b. Find the area of the entire region that lies within both curves.
requals20 sine theta and requals20 cosine theta
Question content area bottom
Part 1
a. Identify all the intersection points. Use 0 for the theta coordinate of the pole if​ it's an intersection point.
  
left parenthesis 0 comma 0 right parenthesis comma left parenthesis 10 StartRoot 2 EndRoot comma StartFraction pi Over 4 EndFraction right parenthesis
​(Type an ordered pair. Type the coordinate for theta in radians. Use a comma to separate answers as needed. Type an exact​ answer, using pi as​ needed.)
Part 2
b. The area of the region that lies within both curves is 
  
enter your response here square units.
​(Type an exact​ answer, using pi as​ needed.)
Use 0 for the coordinate of the pole if it's an intersection point.(Type an ordered pair. Type the coordinate for in radians. Use a comma to separate answers as needed. Type an exact answer, using as needed.)The area of the region that lies within both curves is(Type an exact answer, using as needed.)Use 0 for the coordinate of the pole if it's an intersection point.(Type an ordered pair. Type the coordinate for in radians. Use a comma to separate answers as needed. Type an exact answer, using as needed.)The area of the region that lies within both curves is(Type an exact answer, using as needed.)
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Let's break down the problem step by step:

### Part 1: Intersection Points
You are given two polar curves:
1. \( r = 20 \sin \theta \)
2. \( r = 20 \cos \theta \)

To find the intersection points, we set the two equations equal to each other:
\[
20 \sin \theta = 20 \cos \theta
\]
Dividing both sides by 20 (since it's non-zero), we get:
\[
\sin \theta = \cos \theta
\]

This implies:
\[
\tan \theta = 1
\]
From this, we know that \( \theta = \frac{\pi}{4} \) is the solution in the interval \( [0, 2\pi] \).

Now substitute \( \theta = \frac{\pi}{4} \) into either of the original equations (since both give the same result at intersection points):

For \( r = 20 \sin \theta \):
\[
r = 20 \sin \left( \frac{\pi}{4} \right) = 20 \times \frac{\sqrt{2}}{2} = 10\sqrt{2}
\]

Thus, one of the intersection points is \( \left( 10\sqrt{2}, \frac{\pi}{4} \right) \).

Additionally, both curves pass through the origin \( (0, 0) \), so the second intersection point is \( (0, 0) \).

The intersection points are:
\[
(0, 0), \left( 10\sqrt{2}, \frac{\pi}{4} \right)
\]

### Part 2: Area of the Region
To find the area of the region that lies within both curves, we use the formula for the area in polar coordinates:
\[
A = \frac{1}{2} \int_{\theta_1}^{\theta_2} \min(r_1^2, r_2^2) \, d\theta
\]

We need to integrate the region common to both curves, from \( \theta = 0 \) to \( \theta = \frac{\pi}{4} \), since that's the range where the curves intersect.

The area will be the sum of the two regions, so:
\[
A = 2 \times \frac{1}{2} \int_0^{\frac{\pi}{4}} 20 \sin \theta \cdot 20 \cos \theta \, d\theta
\]
Simplifying this:
\[
A = 20^2 \int_0^{\frac{\pi}{4}} \sin \theta \cos \theta \, d\theta
\]

Using the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \), we get:
\[
A = 200 \int_0^{\frac{\pi}{4}} \frac{\sin(2\theta)}{2} \, d\theta
\]
\[
A = 100 \int_0^{\frac{\pi}{4}} \sin(2\theta) \, d\theta
\]

Now, integrate \( \sin(2\theta) \):
\[
A = 100 \left[ -\frac{\cos(2\theta)}{2} \right]_0^{\frac{\pi}{4}}
\]
\[
A = 100 \times \left( -\frac{\cos\left( \frac{\pi}{2} \right)}{2} + \frac{\cos(0)}{2} \right)
\]
\[
A = 100 \times \left( -\frac{0}{2} + \frac{1}{2} \right) = 100 \times \frac{1}{2} = 50
\]

Thus, the area of the region that lies within both curves is **50 square units**.

---

### Summary:
1. Intersection points: \( (0, 0) \) and \( \left( 10\sqrt{2}, \frac{\pi}{4} \right) \)
2. Area of the region: \( 50 \) square units.

Do you want more details or have any questions?

---

Here are 5 related questions to expand your understanding:
1. How do you find intersection points for more complicated polar curves?
2. What happens if the polar curves don't intersect at the origin?
3. How can you convert polar coordinates to Cartesian coordinates?
4. What is the significance of using the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \)?
5. How would you compute the area if the curves had different angular limits?

**Tip:** Always check for symmetry in polar problems, as it can simplify both intersection and area calculations.

Part 1: a. Find all intersection points of the following curves. b. Find the area of the entire region that lies within both curves. r=20 sin(θ) and r=20 cos(θ)

Learn how to find the intersection points and the area of the region within the polar curves r = 20 sin(θ) and r = 20 cos(θ). This problem involves polar coordinates, trigonometry, and the use of integrals to compute areas. Suitable for students in Grades 11-12.

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