Set up, but do not evaluate, integrals which give the
surface area when y = sin x for 0 ⩽ x ⩽ π is spun
around the line y = −1, and similarly when it is spun
around the line x = π.

To find the surface area generated when the curve \( y = \sin x \) for \( 0 \leq x \leq \pi \) is revolved around different lines, we will use the surface area formula for revolution, which depends on the axis of revolution. We will address the two cases separately.

### Case 1: Spinning around the line \( y = -1 \)

When a curve \( y = f(x) \) is revolved around the horizontal line \( y = -1 \), the formula for the surface area is:

\[
A = 2\pi \int_{a}^{b} (f(x) - (-1)) \sqrt{1 + \left(\frac{df(x)}{dx}\right)^2} \, dx
\]

Here:
- \( f(x) = \sin x \)
- \( f'(x) = \cos x \)
- Limits of integration: \( a = 0 \), \( b = \pi \)

The surface area integral for this case becomes:

\[
A = 2\pi \int_{0}^{\pi} (\sin x + 1) \sqrt{1 + (\cos x)^2} \, dx
\]

This represents the surface area when \( y = \sin x \) is spun around the line \( y = -1 \).

---

### Case 2: Spinning around the line \( x = \pi \)

For revolution around a vertical line \( x = \pi \), the formula is slightly different. The curve will now be treated in terms of \( x = f(y) \). However, in this case, we will adjust the formula for the surface area by revolving around a vertical line, using:

\[
A = 2\pi \int_{y_1}^{y_2} ( \pi - x(y) ) \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy
\]

Here, \( y = \sin x \) so \( x(y) = \arcsin(y) \) and \( \frac{dx}{dy} = \frac{1}{\sqrt{1 - y^2}} \). The limits of integration for \( y \) are \( y_1 = 0 \) and \( y_2 = 1 \) (since \( \sin(0) = 0 \) and \( \sin(\pi) = 1 \)).

Thus, the surface area integral becomes:

\[
A = 2\pi \int_{0}^{1} (\pi - \arcsin(y)) \sqrt{1 + \frac{1}{1 - y^2}} \, dy
\]

This represents the surface area when \( y = \sin x \) is spun around the line \( x = \pi \).

---

### Summary

1. **Revolution around \( y = -1 \)**:
   \[
   A = 2\pi \int_{0}^{\pi} (\sin x + 1) \sqrt{1 + (\cos x)^2} \, dx
   \]

2. **Revolution around \( x = \pi \)**:
   \[
   A = 2\pi \int_{0}^{1} (\pi - \arcsin(y)) \sqrt{1 + \frac{1}{1 - y^2}} \, dy
   \]

Would you like further details on any part of this setup or have any questions?

### Related Questions:
1. How do you derive the surface area formula for a curve revolved around a line?
2. What is the role of the derivative in surface area integrals for revolved curves?
3. How do surface area integrals change when revolving around different axes?
4. What is the geometric interpretation of the surface area generated by revolving a curve?
5. Can this method be applied to other curves and axes of revolution?

### Tip: 
To evaluate these integrals, numerical methods or advanced techniques such as trigonometric identities might be useful if closed forms are not easily obtainable.

Set up, but do not evaluate, integrals which give the surface area when y = sin x for 0 ⩽ x ⩽ π is spun around the line y = −1, and similarly when it is spun around the line x = π.

Set up integrals to calculate the surface area when y = sin x is spun around y = -1 and x = π. This problem covers key calculus concepts like the surface area of revolution, trigonometric functions, and definite integrals. Ideal for students in Calculus II.

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