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!