1. An oceanographer is mapping the ocean floor. Help the oceanographer calculate the volume of water by evaluating the following double integrals over the given rectangular regions $$ R $$ : a) $$ \iint_{R}\left(x y^{2}+2 x y\right) d A $$ where $$ R=\{(x, y) \mid 0 \leq x \leq 4,1 \leq y \leq 3\} $$ (2 marks) b) $$ \iint_{R}\left(\mathrm{e}^{y} \sin x\right) d A $$ where $$ R=\{(x, y) \mid 1 \leq x \leq 2,0 \leq y \leq \pi\} \quad $$ (2 marks)

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**(a) Evaluate $$\iint_R \left(xy^2+2xy\right) dA$$ over $$ R=\{(x,y) \mid 0\leq x \leq 4,\ 1\leq y \leq 3\} $$**

1. Write the double integral as an iterated integral:
   $$
   \iint_R \left(xy^2+2xy\right)dA = \int_{y=1}^{3} \int_{x=0}^{4} \left(xy^2+2xy\right) dx\, dy.
   $$

2. Factor the integrand by taking $$x$$ as a common factor:
   $$
   xy^2+2xy = x\left(y^2+2y\right).
   $$

3. Evaluate the inner integral with respect to $$x$$:
   $$
   \int_{x=0}^{4} x\left(y^2+2y\right)dx = \left(y^2+2y\right) \int_{x=0}^{4} x\, dx.
   $$
   Since
   $$
   \int_{x=0}^{4} x\, dx = \left[\frac{x^2}{2}\right]_0^4 = \frac{4^2}{2} = \frac{16}{2} = 8,
   $$
   the inner integral becomes:
   $$
   8\left(y^2+2y\right).
   $$

4. Now, evaluate the outer integral with respect to $$y$$:
   $$
   \int_{y=1}^{3} 8\left(y^2+2y\right)dy = 8\int_{y=1}^{3} \left(y^2+2y\right)dy.
   $$

5. Compute the antiderivative:
   $$
   \int \left(y^2+2y\right)dy = \frac{y^3}{3}+y^2.
   $$

6. Evaluate the definite integral:
   $$
   \left[\frac{y^3}{3}+y^2\right]_{y=1}^{3} = \left(\frac{3^3}{3}+3^2\right) - \left(\frac{1^3}{3}+1^2\right) = \left(\frac{27}{3}+9\right) - \left(\frac{1}{3}+1\right).
   $$
   Compute the values:
   $$
   \frac{27}{3}+9 = 9+9 = 18,\quad \frac{1}{3}+1 = \frac{4}{3}.
   $$
   So, the result is:
   $$
   18-\frac{4}{3} = \frac{54-4}{3} = \frac{50}{3}.
   $$

7. Multiply by 8:
   $$
   8\cdot\frac{50}{3} = \frac{400}{3}.
   $$

Thus, the value of the double integral is:
$$
\boxed{\frac{400}{3}}.
$$

---

**(b) Evaluate $$\iint_R \left(e^y \sin x\right)dA$$ over $$ R=\{(x, y) \mid 1\leq x \leq 2,\ 0\leq y \leq \pi\} $$**

1. Write the double integral as an iterated integral:
   $$
   \iint_R \left(e^y \sin x\right) dA = \int_{y=0}^{\pi} \int_{x=1}^{2} e^y \sin x\, dx\, dy.
   $$

2. Factor out $$e^y$$ (which is constant with respect to $$x$$):
   $$
   \int_{x=1}^{2} e^y \sin x\, dx = e^y \int_{x=1}^{2} \sin x\, dx.
   $$

3. Evaluate the inner integral with respect to $$x$$:
   $$
   \int_{x=1}^{2} \sin x\, dx = \left[-\cos x\right]_{x=1}^{2} = \big[-\cos 2\big] - \big[-\cos 1\big] = \cos 1-\cos 2.
   $$

4. Substitute back into the integral:
   $$
   \iint_R e^y \sin x \, dA = \left(\cos 1-\cos 2\right) \int_{y=0}^{\pi} e^y\, dy.
   $$

5. Evaluate the integral with respect to $$y$$:
   $$
   \int_{y=0}^{\pi} e^y dy = \left[e^y\right]_{y=0}^{\pi} = e^{\pi}-e^0 = e^{\pi}-1.
   $$

6. Therefore, the value of the double integral is:
   $$
   \left(\cos 1-\cos 2\right)\left(e^{\pi}-1\right).
   $$

Thus, the final answer for part (b) is:
$$
\boxed{\left(\cos 1-\cos 2\right)\left(e^{\pi}-1\right)}.
$$
**(a) The volume is $$ \frac{400}{3} $$.** **(b) The volume is $$ (\cos 1 - \cos 2)(e^{\pi} - 1) $$.

An oceanographer is mapping the ocean floor. Help the oceanographer calculate the
volume of water by evaluating the following double integrals over the given rectangular
regions :
a) where (2 marks)
b) where (2 marks)

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An oceanographer is mapping the ocean floor. Help the oceanographer calculate the volume of water by evaluating the following double integrals over the given rectangular regions : a) where (2 marks) b) where (2 marks)