1. **Problem:** Evaluate the double integral $$\int_0^1 \int_0^2 4xy \, dx \, dy$$. 2. **Formula and rules:** For double integrals, integrate with respect to the inner variable first, then the outer. 3. **Step 1:** Integrate with respect to $x$: $$\int_0^2 4xy \, dx = 4y \int_0^2 x \, dx = 4y \left[ \frac{x^2}{2} \right]_0^2 = 4y \times 2 = 8y$$ 4. **Step 2:** Integrate with respect to $y$: $$\int_0^1 8y \, dy = 8 \left[ \frac{y^2}{2} \right]_0^1 = 8 \times \frac{1}{2} = 4$$ 5. **Answer:** The value of the integral is $4$. --- 2. **Problem:** Evaluate $$\int_1^b \int_1^a \frac{1}{xy} \, dx \, dy$$. 3. **Step 1:** Integrate with respect to $x$: $$\int_1^a \frac{1}{xy} \, dx = \frac{1}{y} \int_1^a \frac{1}{x} \, dx = \frac{1}{y} [\ln x]_1^a = \frac{\ln a}{y}$$ 4. **Step 2:** Integrate with respect to $y$: $$\int_1^b \frac{\ln a}{y} \, dy = \ln a \int_1^b \frac{1}{y} \, dy = \ln a [\ln y]_1^b = \ln a \ln b$$ 5. **Answer:** The value of the integral is $\ln a \ln b$. --- 3. **Problem:** Evaluate $$\int_0^1 \int_0^x 1 \, dy \, dx$$. 4. **Step 1:** Integrate with respect to $y$: $$\int_0^x 1 \, dy = [y]_0^x = x$$ 5. **Step 2:** Integrate with respect to $x$: $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$$ 6. **Answer:** The value of the integral is $\frac{1}{2}$. --- 4. **Problem:** Evaluate $$\int_0^\pi \int_0^{\sin \theta} r \, dr \, d\theta$$. 5. **Step 1:** Integrate with respect to $r$: $$\int_0^{\sin \theta} r \, dr = \left[ \frac{r^2}{2} \right]_0^{\sin \theta} = \frac{\sin^2 \theta}{2}$$ 6. **Step 2:** Integrate with respect to $\theta$: $$\int_0^\pi \frac{\sin^2 \theta}{2} \, d\theta = \frac{1}{2} \int_0^\pi \sin^2 \theta \, d\theta$$ 7. **Recall:** $$\int_0^\pi \sin^2 \theta \, d\theta = \frac{\pi}{2}$$ 8. **Therefore:** $$\frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}$$ 9. **Answer:** The value of the integral is $\frac{\pi}{4}$. --- 5. **Problem:** Evaluate $$\int_0^1 \int_0^2 \int_0^3 xyz \, dx \, dy \, dz$$. 6. **Step 1:** Integrate with respect to $x$: $$\int_0^3 xyz \, dx = yz \int_0^3 x \, dx = yz \left[ \frac{x^2}{2} \right]_0^3 = yz \times \frac{9}{2} = \frac{9}{2} yz$$ 7. **Step 2:** Integrate with respect to $y$: $$\int_0^2 \frac{9}{2} yz \, dy = \frac{9}{2} z \int_0^2 y \, dy = \frac{9}{2} z \left[ \frac{y^2}{2} \right]_0^2 = \frac{9}{2} z \times 2 = 9z$$ 8. **Step 3:** Integrate with respect to $z$: $$\int_0^1 9z \, dz = 9 \left[ \frac{z^2}{2} \right]_0^1 = 9 \times \frac{1}{2} = \frac{9}{2}$$ 9. **Answer:** The value of the integral is $\frac{9}{2}$. --- 6. **Problem:** Evaluate $$\int_0^1 \int_0^z \int_0^{y+z} dz \, dy \, dx$$. 7. **Step 1:** Integrate with respect to $z$ (inner integral): $$\int_0^{y+z} dz = [z]_0^{y+z} = y + z$$ 8. **Step 2:** Integrate with respect to $y$: $$\int_0^z (y + z) \, dy = \int_0^z y \, dy + \int_0^z z \, dy = \left[ \frac{y^2}{2} \right]_0^z + z[y]_0^z = \frac{z^2}{2} + z^2 = \frac{3z^2}{2}$$ 9. **Step 3:** Integrate with respect to $x$: $$\int_0^1 \frac{3z^2}{2} \, dx = \frac{3z^2}{2} [x]_0^1 = \frac{3z^2}{2}$$ 10. **Note:** The variable $z$ is a dummy variable of integration and should be integrated over its limits. However, the problem as stated has an ambiguity because the outermost integral is with respect to $x$ but the integrand depends on $z$. Assuming the problem meant $$\int_0^1 \int_0^z \int_0^{y+z} dz \, dy \, dz$$ (outer integral over $z$), then: 11. **Step 4:** Integrate with respect to $z$: $$\int_0^1 \frac{3z^2}{2} \, dz = \frac{3}{2} \left[ \frac{z^3}{3} \right]_0^1 = \frac{3}{2} \times \frac{1}{3} = \frac{1}{2}$$ 12. **Answer:** The value of the integral is $\frac{1}{2}$. --- **Summary:** 1. $4$ 2. $\ln a \ln b$ 3. $\frac{1}{2}$ 4. $\frac{\pi}{4}$ 5. $\frac{9}{2}$ 6. $\frac{1}{2}$