find the area of the region enclosed by curves x=5siny and x=(10y)/pi using integrals

To find the area of the region enclosed by the curves \( x = 5\sin(y) \) and \( x = \frac{10y}{\pi} \), we need to follow these steps:

### Step 1: Find the points of intersection
First, we need to determine the points where the two curves intersect. We do this by setting the two expressions for \( x \) equal to each other:

\[
5\sin(y) = \frac{10y}{\pi}
\]

Simplifying this:

\[
\sin(y) = \frac{2y}{\pi}
\]

To solve for \( y \), we can examine the intersections graphically or numerically, but for now, let's note that this equation defines the bounds for \( y \).

### Step 2: Determine the integral bounds
The solutions to \( \sin(y) = \frac{2y}{\pi} \) give us the values of \( y \) where the curves intersect. We can find these intersections by recognizing that \( y = 0 \) is one intersection point (since \( \sin(0) = 0 \) and \( \frac{2 \times 0}{\pi} = 0 \)). 

The second intersection occurs at a non-trivial value of \( y \), which we'll denote as \( y_1 \). We can estimate \( y_1 \) by solving the equation numerically or observing graph behavior, but for now, let's represent it generally and assume that the curve intersection occurs within a known range.

### Step 3: Set up the integral for the area
The area \( A \) between two curves is given by the integral:

\[
A = \int_{y_0}^{y_1} \left[ f(y) - g(y) \right] \, dy
\]

Where:
- \( f(y) \) is the upper curve
- \( g(y) \) is the lower curve
- \( y_0 \) and \( y_1 \) are the intersection points

Here, \( f(y) = \frac{10y}{\pi} \) and \( g(y) = 5\sin(y) \), so:

\[
A = \int_{y_0}^{y_1} \left[ \frac{10y}{\pi} - 5\sin(y) \right] \, dy
\]

### Step 4: Evaluate the integral
Assuming \( y_0 = 0 \) and finding \( y_1 \) from \( \sin(y) = \frac{2y}{\pi} \), we can write the integral as:

\[
A = \int_{0}^{y_1} \left( \frac{10y}{\pi} - 5\sin(y) \right) \, dy
\]

Let's compute this integral in parts:

\[
\int \frac{10y}{\pi} \, dy = \frac{10}{\pi} \cdot \frac{y^2}{2} = \frac{5y^2}{\pi}
\]

\[
\int 5\sin(y) \, dy = -5\cos(y)
\]

Thus:

\[
A = \left[ \frac{5y^2}{\pi} + 5\cos(y) \right]_{0}^{y_1}
\]

Substituting the limits:

\[
A = \frac{5y_1^2}{\pi} + 5\cos(y_1) - \left( 0 + 5\cos(0) \right)
\]

\[
A = \frac{5y_1^2}{\pi} + 5\cos(y_1) - 5
\]

### Step 5: Approximate \( y_1 \)
\( y_1 \) is the solution to \( \sin(y) = \frac{2y}{\pi} \) that lies in \( (0, \frac{\pi}{2}) \). A numerical approach or estimation might be necessary to find \( y_1 \).

For a rough estimate, \( y_1 \approx 1.4 \) based on \( \sin(y) \approx \frac{2y}{\pi} \) near this value.

### Final Expression:
With \( y_1 \approx 1.4 \):

\[
A \approx \frac{5 \times (1.4)^2}{\pi} + 5\cos(1.4) - 5
\]

This results in an approximate value for the area. For precise results, you would calculate \( y_1 \) more accurately using numerical methods.

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Would you like details on numerical methods to find \( y_1 \), further explanations, or do you have any other questions?

---

1. How do you solve \( \sin(y) = \frac{2y}{\pi} \) numerically?
2. What are the exact steps to evaluate the integral once \( y_1 \) is known?
3. Can the integral be simplified or solved symbolically for specific values of \( y_1 \)?
4. How does changing the parameter \( \pi \) affect the area?
5. What numerical tools can be used to evaluate complex integrals like this one?

**Tip:** Checking the behavior of functions graphically can help identify approximate solutions before applying numerical methods.

Find the area of the region enclosed by curves x=5siny and x=(10y)/pi using integrals.

Learn how to find the area of the region enclosed by the curves x=5siny and x=(10y)/pi using integrals. This problem explores key calculus concepts such as the area between curves and integration, suitable for students in Grades 11-12.

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