https://file.aiquickdraw.com/mathbot/file/172734777059351vsgaif.jpg

The given image shows a triple integral of the following form:

\[
\int_0^{\frac{\pi}{2}} \int_0^z \int_0^y \sin(x + y + z) \, dx \, dy \, dz
\]

We need to evaluate this integral step-by-step.

### Step-by-step solution:

1. **Inner integral (with respect to \(x\)):**
   The first integral is with respect to \(x\), keeping \(y\) and \(z\) as constants.
   \[
   \int_0^y \sin(x + y + z) \, dx
   \]
   To integrate this, let’s use the substitution \( u = x + y + z \), so \( du = dx \). When \(x = 0\), \( u = y + z \), and when \(x = y\), \( u = 2y + z \). The integral becomes:
   \[
   \int_{y+z}^{2y+z} \sin(u) \, du
   \]
   The integral of \( \sin(u) \) is \( -\cos(u) \), so this becomes:
   \[
   -\cos(u) \Big|_{y+z}^{2y+z} = -\cos(2y+z) + \cos(y+z)
   \]
   Thus, the result of the inner integral is:
   \[
   \cos(y+z) - \cos(2y+z)
   \]

2. **Second integral (with respect to \(y\)):**
   Now we need to integrate this result with respect to \(y\):
   \[
   \int_0^z [\cos(y+z) - \cos(2y+z)] \, dy
   \]
   For each term:
   - \( \int_0^z \cos(y+z) \, dy \) can be simplified using the substitution \( u = y + z \). This leads to \( \sin(y + z) \), evaluated from \(y = 0\) to \(y = z\), yielding:
     \[
     \sin(2z) - \sin(z)
     \]
   - \( \int_0^z \cos(2y+z) \, dy \) gives a similar result. Substituting \( u = 2y + z \) results in \( \frac{1}{2} \sin(2y+z) \), evaluated from \(y = 0\) to \(y = z\), yielding:
     \[
     \frac{1}{2} [\sin(3z) - \sin(z)]
     \]

   Combining these, we get:
   \[
   (\sin(2z) - \sin(z)) - \frac{1}{2} [\sin(3z) - \sin(z)]
   \]

3. **Final integral (with respect to \(z\)):**
   Finally, integrate this result with respect to \(z\):
   \[
   \int_0^{\frac{\pi}{2}} \left( \sin(2z) - \sin(z) - \frac{1}{2} [\sin(3z) - \sin(z)] \right) dz
   \]
   Breaking this up:
   - \( \int_0^{\frac{\pi}{2}} \sin(2z) \, dz = -\frac{1}{2} \cos(2z) \Big|_0^{\frac{\pi}{2}} = 1 \)
   - \( \int_0^{\frac{\pi}{2}} \sin(z) \, dz = -\cos(z) \Big|_0^{\frac{\pi}{2}} = 2 \)
   - \( \int_0^{\frac{\pi}{2}} \sin(3z) \, dz = -\frac{1}{3} \cos(3z) \Big|_0^{\frac{\pi}{2}} = \frac{1}{3} \)
   - \( \int_0^{\frac{\pi}{2}} \sin(z) \, dz = 2 \), so \( \frac{1}{2} \times 2 = 1 \).

   Combining all these:
   \[
   1 - 2 - \frac{1}{2} \times \left( \frac{1}{3} - 2 \right) = 1 - 2 - \frac{1}{2} \times \left( -\frac{5}{3} \right)
   \]
   Simplifying:
   \[
   -1 + \frac{5}{6} = -\frac{1}{6}
   \]

Thus, the value of the triple integral is:
\[
\boxed{-\frac{1}{6}}
\]

Would you like further details on any step or clarification? Here are 5 related questions to practice:

1. How do you evaluate a triple integral in cylindrical coordinates?
2. What is the Jacobian determinant for a change of variables in multiple integrals?
3. How can you change the order of integration in a triple integral?
4. What is the physical interpretation of a triple integral in real-world applications?
5. Can you evaluate a triple integral using spherical coordinates for a given function?

**Tip:** Always check the limits of integration carefully, especially when changing variables or switching the order of integration.

Evaluate the triple integral \(\int_0^{\frac{\pi}{2}} \int_0^z \int_0^y \sin(x + y + z) \, dx \, dy \, dz\).

Learn how to evaluate the triple integral \(\int_0^{\frac{\pi}{2}} \int_0^z \int_0^y \sin(x + y + z) \, dx \, dy \, dz\) using substitution and trigonometric integration. This detailed step-by-step solution is ideal for students in advanced calculus.

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