Consider the region between y=sin(x) and y=cos(x) on the interval [0, pi/4]. Find the volume of the solid obtained by rotating the region about the horizontal line y = -1.


To find the volume of the solid obtained by rotating the region between \( y = \sin(x) \) and \( y = \cos(x) \) on the interval \( [0, \frac{\pi}{4}] \) around the horizontal line \( y = -1 \), we will use the method of washers (or disks with holes).

### Step 1: General formula for volume
The volume of a solid obtained by rotating a region about a horizontal line using the washer method is given by:

\[
V = \pi \int_a^b \left( R_{\text{outer}}^2(x) - R_{\text{inner}}^2(x) \right) \, dx
\]
where:
- \( R_{\text{outer}}(x) \) is the distance from the rotation axis to the outer curve (in this case, the curve further from \( y = -1 \)),
- \( R_{\text{inner}}(x) \) is the distance from the rotation axis to the inner curve (the curve closer to \( y = -1 \)),
- \( a \) and \( b \) are the bounds of the region, here \( a = 0 \) and \( b = \frac{\pi}{4} \).

### Step 2: Determine the radii
- The outer curve is \( y = \cos(x) \), so the distance from the line \( y = -1 \) to \( y = \cos(x) \) is:
\[
R_{\text{outer}}(x) = \cos(x) - (-1) = \cos(x) + 1
\]
- The inner curve is \( y = \sin(x) \), so the distance from the line \( y = -1 \) to \( y = \sin(x) \) is:
\[
R_{\text{inner}}(x) = \sin(x) - (-1) = \sin(x) + 1
\]

### Step 3: Set up the integral
Now we plug these expressions into the volume formula:

\[
V = \pi \int_0^{\frac{\pi}{4}} \left[ (\cos(x) + 1)^2 - (\sin(x) + 1)^2 \right] \, dx
\]

### Step 4: Simplify the integrand
First, expand the squares:

\[
(\cos(x) + 1)^2 = \cos^2(x) + 2\cos(x) + 1
\]
\[
(\sin(x) + 1)^2 = \sin^2(x) + 2\sin(x) + 1
\]

Now substitute these into the integral:

\[
V = \pi \int_0^{\frac{\pi}{4}} \left[ (\cos^2(x) + 2\cos(x) + 1) - (\sin^2(x) + 2\sin(x) + 1) \right] \, dx
\]

Simplify the integrand:

\[
V = \pi \int_0^{\frac{\pi}{4}} \left[ \cos^2(x) - \sin^2(x) + 2\cos(x) - 2\sin(x) \right] \, dx
\]

Using the identity \( \cos^2(x) - \sin^2(x) = \cos(2x) \), we have:

\[
V = \pi \int_0^{\frac{\pi}{4}} \left[ \cos(2x) + 2\cos(x) - 2\sin(x) \right] \, dx
\]

### Step 5: Evaluate the integral
We can now integrate term by term:
\[
\int_0^{\frac{\pi}{4}} \cos(2x) \, dx = \frac{1}{2} \sin(2x) \Big|_0^{\frac{\pi}{4}} = \frac{1}{2} \sin\left( \frac{\pi}{2} \right) - \frac{1}{2} \sin(0) = \frac{1}{2}
\]
\[
\int_0^{\frac{\pi}{4}} 2\cos(x) \, dx = 2\sin(x) \Big|_0^{\frac{\pi}{4}} = 2\sin\left( \frac{\pi}{4} \right) - 2\sin(0) = 2 \times \frac{\sqrt{2}}{2} = \sqrt{2}
\]
\[
\int_0^{\frac{\pi}{4}} (-2\sin(x)) \, dx = 2\cos(x) \Big|_0^{\frac{\pi}{4}} = 2\cos\left( \frac{\pi}{4} \right) - 2\cos(0) = 2 \times \frac{\sqrt{2}}{2} - 2 = \sqrt{2} - 2
\]

### Step 6: Combine results
Now sum up the individual integrals:

\[
V = \pi \left( \frac{1}{2} + \sqrt{2} + (\sqrt{2} - 2) \right) = \pi \left( \frac{1}{2} + 2\sqrt{2} - 2 \right)
\]

Thus, the volume is:

\[
V = \pi \left( 2\sqrt{2} - \frac{3}{2} \right)
\]

This is the exact volume of the solid.

Would you like more details or any clarifications?

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Here are 5 related questions to explore:

1. What if the rotation was around \( y = 1 \) instead of \( y = -1 \)?
2. How does the method of cylindrical shells apply to this problem?
3. How would the volume change if we rotated the region around the \( x \)-axis?
4. What happens if we extend the interval to \( [0, \frac{\pi}{2}] \)?
5. Can we solve this using numerical methods for approximate integration?

**Tip**: Always double-check your limits of integration, especially when dealing with trigonometric functions and rotation axes!

Consider the region between y = sin(x) and y = cos(x) on the interval [0, pi/4]. Find the volume of the solid obtained by rotating the region about the horizontal line y = -1.

Explore the step-by-step solution to finding the volume of the solid obtained by rotating the region between y = sin(x) and y = cos(x) on the interval [0, pi/4] around the horizontal line y = -1. This calculus problem involves using the washer method and applying trigonometric identities.

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