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.