Here is a transcription of the math problems from the image: --- **1. Evaluate each of the following double integrals.** (a) \(\int_0^1 \int_1^4 \left(\frac{2x}{y} - \frac{1}{\sqrt{y}} \right) dy \, dx\) (b) \(\int_3^5 \int_0^x \frac{x \ln y}{y} \, dx \, dy\) (c) \(\int_0^1 \int_0^{6} xy \cos (3x) \, dx \, dy\) --- **2. (a) Find the volume contained beneath \(z = 3x + 2y\) and above the \(xy\)-plane over the rectangle \(0 \leq x \leq 1\), \(0 \leq y \leq 2\).** (b) Find the volume contained between the surfaces \(z = 3x + 2y\) and \(z = -x^2 - y^2\) over the same rectangle. --- **3. Let \(D\) be the region in the \(xy\)-plane bounded by \(x = 0\), \(y = 1-x\), and \(y = 2-2x\).** (a) Sketch the region \(D\), and say whether it is type \(x\), type \(y\), both, or neither. (b) Set up, but do not evaluate, one double integral to integrate \(f(x, y) = xy\) over the region \(D\). --- **4. Let \(D\) be the region in the first quadrant of the \(xy\)-plane given by \(1 \leq x^2 + y^2 \leq 4\).** (a) Sketch the region \(D\), and say whether it is type \(x\), type \(y\), both, or neither. (b) Set up, but do not evaluate, a double integral or sum of double integrals to integrate \(f(x, y) = xy\) over the region \(D\). --- **5. Consider the double integral \(\int_0^1 \int_y^1 e^{x^2} dx \, dy\).** (a) Sketch the region being integrated over. (b) Rewrite the integral in the opposite order of integration, then evaluate. Notice that we cannot evaluate the integral by hand when it is written in the original order. --- If you need solutions, explanations, or sketches for any of these problems, please let me know which one you'd like to tackle!