This worksheet presents four distinct double integral problems for calculus students to solve.

Problem: Evaluate the given double integrals.

We will solve each problem step by step.

---

#### Problem 1:
$$
\iint_{1}^{3} \int_{2}^{4} 9x^3 y^2 \, dy \, dx
$$

Step 1: Inner integral with respect to \( y \).
The inner integral is:
$$
\int_{2}^{4} 9x^3 y^2 \, dy
$$
Here, \( x \) is treated as a constant. Factor out \( 9x^3 \):
$$
\int_{2}^{4} 9x^3 y^2 \, dy = 9x^3 \int_{2}^{4} y^2 \, dy
$$
Now, integrate \( y^2 \):
$$
\int y^2 \, dy = \frac{y^3}{3}
$$
Evaluate this from \( y = 2 \) to \( y = 4 \):
$$
\int_{2}^{4} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{2}^{4} = \frac{4^3}{3} - \frac{2^3}{3} = \frac{64}{3} - \frac{8}{3} = \frac{56}{3}
$$
Thus, the inner integral becomes:
$$
9x^3 \int_{2}^{4} y^2 \, dy = 9x^3 \cdot \frac{56}{3} = 168x^3
$$

Step 2: Outer integral with respect to \( x \).
Now, integrate \( 168x^3 \) with respect to \( x \) from \( x = 1 \) to \( x = 3 \):
$$
\int_{1}^{3} 168x^3 \, dx
$$
Factor out the constant \( 168 \):
$$
\int_{1}^{3} 168x^3 \, dx = 168 \int_{1}^{3} x^3 \, dx
$$
Integrate \( x^3 \):
$$
\int x^3 \, dx = \frac{x^4}{4}
$$
Evaluate this from \( x = 1 \) to \( x = 3 \):
$$
\int_{1}^{3} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{1}^{3} = \frac{3^4}{4} - \frac{1^4}{4} = \frac{81}{4} - \frac{1}{4} = \frac{80}{4} = 20
$$
Thus, the outer integral becomes:
$$
168 \int_{1}^{3} x^3 \, dx = 168 \cdot 20 = 3360
$$

Final Answer for Problem 1:
$$
\boxed{3360}
$$

---

#### Problem 2:
$$
\int_{11}^{14} \int_{7}^{10} (x^2 + 4y) \, dy \, dx
$$

Step 1: Inner integral with respect to \( y \).
The inner integral is:
$$
\int_{7}^{10} (x^2 + 4y) \, dy
$$
Here, \( x \) is treated as a constant. Split the integral:
$$
\int_{7}^{10} (x^2 + 4y) \, dy = \int_{7}^{10} x^2 \, dy + \int_{7}^{10} 4y \, dy
$$
For the first term, \( x^2 \) is a constant:
$$
\int_{7}^{10} x^2 \, dy = x^2 \int_{7}^{10} 1 \, dy = x^2 \left[ y \right]_{7}^{10} = x^2 (10 - 7) = 3x^2
$$
For the second term:
$$
\int_{7}^{10} 4y \, dy = 4 \int_{7}^{10} y \, dy = 4 \left[ \frac{y^2}{2} \right]_{7}^{10} = 4 \left( \frac{10^2}{2} - \frac{7^2}{2} \right) = 4 \left( \frac{100}{2} - \frac{49}{2} \right) = 4 \left( \frac{51}{2} \right) = 102
$$
Combine the results:
$$
\int_{7}^{10} (x^2 + 4y) \, dy = 3x^2 + 102
$$

Step 2: Outer integral with respect to \( x \).
Now, integrate \( 3x^2 + 102 \) with respect to \( x \) from \( x = 11 \) to \( x = 14 \):
$$
\int_{11}^{14} (3x^2 + 102) \, dx
$$
Split the integral:
$$
\int_{11}^{14} (3x^2 + 102) \, dx = \int_{11}^{14} 3x^2 \, dx + \int_{11}^{14} 102 \, dx
$$
For the first term:
$$
\int_{11}^{14} 3x^2 \, dx = 3 \int_{11}^{14} x^2 \, dx = 3 \left[ \frac{x^3}{3} \right]_{11}^{14} = \left[ x^3 \right]_{11}^{14} = 14^3 - 11^3 = 2744 - 1331 = 1413
$$
For the second term:
$$
\int_{11}^{14} 102 \, dx = 102 \int_{11}^{14} 1 \, dx = 102 \left[ x \right]_{11}^{14} = 102 (14 - 11) = 102 \cdot 3 = 306
$$
Combine the results:
$$
\int_{11}^{14} (3x^2 + 102) \, dx = 1413 + 306 = 1719
$$

Final Answer for Problem 2:
$$
\boxed{1719}
$$

---

#### Problem 3:
$$
\int_{0}^{\pi/2} \int_{0}^{\pi/2} x \cos(xy) \, dy \, dx
$$

Step 1: Inner integral with respect to \( y \).
The inner integral is:
$$
\int_{0}^{\pi/2} x \cos(xy) \, dy
$$
Here, \( x \) is treated as a constant. Factor out \( x \):
$$
\int_{0}^{\pi/2} x \cos(xy) \, dy = x \int_{0}^{\pi/2} \cos(xy) \, dy
$$
To integrate \( \cos(xy) \), use the substitution \( u = xy \). Then \( du = x \, dy \), and when \( y = 0 \), \( u = 0 \); when \( y = \pi/2 \), \( u = x(\pi/2) \). The integral becomes:
$$
\int_{0}^{\pi/2} \cos(xy) \, dy = \int_{0}^{x\pi/2} \cos(u) \cdot \frac{du}{x} = \frac{1}{x} \int_{0}^{x\pi/2} \cos(u) \, du
$$
Integrate \( \cos(u) \):
$$
\int \cos(u) \, du = \sin(u)
$$
Evaluate this from \( u = 0 \) to \( u = x\pi/2 \):
$$
\int_{0}^{x\pi/2} \cos(u) \, du = \left[ \sin(u) \right]_{0}^{x\pi/2} = \sin\left(\frac{x\pi}{2}\right) - \sin(0) = \sin\left(\frac{x\pi}{2}\right)
$$
Thus, the inner integral becomes:
$$
x \int_{0}^{\pi/2} \cos(xy) \, dy = x \cdot \frac{1}{x} \sin\left(\frac{x\pi}{2}\right) = \sin\left(\frac{x\pi}{2}\right)
$$

Step 2: Outer integral with respect to \( x \).
Now, integrate \( \sin\left(\frac{x\pi}{2}\right) \) with respect to \( x \) from \( x = 0 \) to \( x = \pi/2 \):
$$
\int_{0}^{\pi/2} \sin\left(\frac{x\pi}{2}\right) \, dx
$$
Use the substitution \( v = \frac{x\pi}{2} \). Then \( dv = \frac{\pi}{2} \, dx \), and when \( x = 0 \), \( v = 0 \); when \( x = \pi/2 \), \( v = \pi/2 \). The integral becomes:
$$
\int_{0}^{\pi/2} \sin\left(\frac{x\pi}{2}\right) \, dx = \int_{0}^{\pi/2} \sin(v) \cdot \frac{2}{\pi} \, dv = \frac{2}{\pi} \int_{0}^{\pi/2} \sin(v) \, dv
$$
Integrate \( \sin(v) \):
$$
\int \sin(v) \, dv = -\cos(v)
$$
Evaluate this from \( v = 0 \) to \( v = \pi/2 \):
$$
\int_{0}^{\pi/2} \sin(v) \, dv = \left[ -\cos(v) \right]_{0}^{\pi/2} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = -0 + 1 = 1
$$
Thus, the outer integral becomes:
$$
\frac{2}{\pi} \int_{0}^{\pi/2} \sin(v) \, dv = \frac{2}{\pi} \cdot 1 = \frac{2}{\pi}
$$

Final Answer for Problem 3:
$$
\boxed{\frac{2}{\pi}}
$$

---

#### Problem 4:
$$
\int_{1}^{4} \int_{0}^{\sqrt{y}} e^{x/\sqrt{y}} \, dy \, dx
$$

Step 1: Inner integral with respect to \( x \).
The inner integral is:
$$
\int_{0}^{\sqrt{y}} e^{x/\sqrt{y}} \, dx
$$
Here, \( y \) is treated as a constant. Let \( k = \frac{1}{\sqrt{y}} \). Then the integral becomes:
$$
\int_{0}^{\sqrt{y}} e^{kx} \, dx
$$
Integrate \( e^{kx} \):
$$
\int e^{kx} \, dx = \frac{e^{kx}}{k}
$$
Evaluate this from \( x = 0 \) to \( x = \sqrt{y} \):
$$
\int_{0}^{\sqrt{y}} e^{kx} \, dx = \left[ \frac{e^{kx}}{k} \right]_{0}^{\sqrt{y}} = \frac{e^{k\sqrt{y}}}{k} - \frac{e^{k \cdot 0}}{k} = \frac{e^{k\sqrt{y}}}{k} - \frac{1}{k}
$$
Substitute back \( k = \frac{1}{\sqrt{y}} \):
$$
\frac{e^{k\sqrt{y}}}{k} - \frac{1}{k} = \frac{e^{(1/\sqrt{y})\sqrt{y}}}{1/\sqrt{y}} - \frac{1}{1/\sqrt{y}} = \frac{e^1}{1/\sqrt{y}} - \frac{1}{1/\sqrt{y}} = \sqrt{y}e - \sqrt{y}
$$
Thus, the inner integral becomes:
$$
\int_{0}^{\sqrt{y}} e^{x/\sqrt{y}} \, dx = \sqrt{y}(e - 1)
$$

Step 2: Outer integral with respect to \( y \).
Now, integrate \( \sqrt{y}(e - 1) \) with respect to \( y \) from \( y = 1 \) to \( y = 4 \):
$$
\int_{1}^{4} \sqrt{y}(e - 1) \, dy
$$
Factor out the constant \( e - 1 \):
$$
\int_{1}^{4} \sqrt{y}(e - 1) \, dy = (e - 1) \int_{1}^{4} \sqrt{y} \, dy
$$
Rewrite \( \sqrt{y} \) as \( y^{1/2} \):
$$
\int_{1}^{4} \sqrt{y} \, dy = \int_{1}^{4} y^{1/2} \, dy
$$
Integrate \( y^{1/2} \):
$$
\int y^{1/2} \, dy = \frac{y^{3/2}}{3/2} = \frac{2}{3} y^{3/2}
$$
Evaluate this from \( y = 1 \) to \( y = 4 \):
$$
\int_{1}^{4} y^{1/2} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_{1}^{4} = \frac{2}{3} (4^{3/2}) - \frac{2}{3} (1^{3/2}) = \frac{2}{3} (8) - \frac{2}{3} (1) = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}
$$
Thus, the outer integral becomes:
$$
(e - 1) \int_{1}^{4} \sqrt{y} \, dy = (e - 1) \cdot \frac{14}{3} = \frac{14(e - 1)}{3}
$$

Final Answer for Problem 4:
$$
\boxed{\frac{14(e - 1)}{3}}
$$

---

Final Answers:

1. \( \boxed{3360} \)
2. \( \boxed{1719} \)
3. \( \boxed{\frac{2}{\pi}} \)
4. \( \boxed{\frac{14(e - 1)}{3}} \)