Let's carefully rewrite the given integral in the order \(dy\, dx\, dz\). ### **Step 1: Write the given integrals** \[ I = \int_0^1 \int_0^z \int_0^{1-z} dx\, dy\, dz + \int_0^1 \int_z^1 \int_0^{1-y} dx\, dy\, dz \] ### **Step 2: Analyze the regions** #### **First integral** \[ \int_0^1 \int_0^z \int_0^{1-z} dx\, dy\, dz \] - \(z\) goes from \(0\) to \(1\) - \(y\) goes from \(0\) to \(z\) - \(x\) goes from \(0\) to \(1-z\) #### **Second integral** \[ \int_0^1 \int_z^1 \int_0^{1-y} dx\, dy\, dz \] - \(z\) goes from \(0\) to \(1\) - \(y\) goes from \(z\) to \(1\) - \(x\) goes from \(0\) to \(1-y\) ### **Step 3: Combine the regions in the \(y-z\) plane** Let’s describe the full region in the \(y, z\) plane: - For the first integral: \(0 \leq z \leq 1\), \(0 \leq y \leq z\) - For the second: \(0 \leq z \leq 1\), \(z \leq y \leq 1\) So, combined: \(0 \leq z \leq 1\), \(0 \leq y \leq 1\) Now, for each region: - If \(0 \leq y \leq z\): \(x\) goes from \(0\) to \(1-z\) - If \(z \leq y \leq 1\): \(x\) goes from \(0\) to \(1-y\) ### **Step 4: Express in the order \(dy\, dx\, dz\)** Let's split the integration over \(y\) at \(y=z\): \[ I = \int_{z=0}^1 \left[ \int_{y=0}^z \int_{x=0}^{1-z} dx\, dy + \int_{y=z}^1 \int_{x=0}^{1-y} dx\, dy \right] dz \] Now, switch the order to \(dy\, dx\, dz\). For each region: #### **For \(0 \leq y \leq z\):** - \(z\) goes from \(y\) to \(1\) - \(y\) goes from \(0\) to \(1\) - \(x\) goes from \(0\) to \(1-z\) #### **For \(z \leq y \leq 1\):** - \(z\) goes from \(0\) to \(y\) - \(y\) goes from \(0\) to \(1\) - \(x\) goes from \(0\) to \(1-y\) #### Let's write both: #### **Region 1** (\(0 \leq y \leq z \leq 1\)): \[ \int_{y=0}^1 \int_{z=y}^1 \int_{x=0}^{1-z} dx\, dz\, dy \] #### **Region 2** (\(0 \leq z \leq y \leq 1\)): \[ \int_{y=0}^1 \int_{z=0}^y \int_{x=0}^{1-y} dx\, dz\, dy \] ### **Step 5: Combine and write the final answer** \[ \boxed{ I = \int_0^1 \int_y^1 \int_0^{1-z} dx\, dz\, dy + \int_0^1 \int_0^y \int_0^{1-y} dx\, dz\, dy } \] --- #### **Summary** - We rewrote the expression in the order \(dy\, dx\, dz\). - The new bounds are: - For the first term: \(0 \leq y \leq 1\), \(y \leq z \leq 1\), \(0 \leq x \leq 1-z\) - For the second term: \(0 \leq y \leq 1\), \(0 \leq z \leq y\), \(0 \leq x \leq 1-y\)