Solution:

We can set up this integral two ways. The left-hand approach, \(\displaystyle \int_1^2 \int_0^\pi y\sin(xy)\,dy\,dx\), would require integration by parts and is more difficult.

Instead, consider integrating with respect to \(\displaystyle x\) first:

\[\iint_R y\sin(xy)\,dA = \int_0^\pi \int_1^2 y\sin(xy)\,dx\,dy\]

Evaluating the inner integral with respect to \(\displaystyle x\). Since \(\displaystyle \int y\sin(xy)\,dx = -\cos(xy)\) (treating \(\displaystyle y\) as constant):

\[= \int_0^\pi \left[-\cos(xy)\right]_{x=1}^{x=2}\,dy = \int_0^\pi \left[-\cos(2y) + \cos(y)\right]\,dy\]

Now integrating with respect to \(\displaystyle y\):

\[= \left[-\frac{\sin(2y)}{2} + \sin(y)\right]_0^\pi\] \[= \left(-\frac{\sin(2\pi)}{2} + \sin(\pi)\right) - \left(-\frac{\sin(0)}{2} + \sin(0)\right) = 0\]

Solution:

This is a Type I region. Fix \(\displaystyle x\): the bounds on \(\displaystyle y\) are \(\displaystyle 2x^2 \leq y \leq 1 + x^2\).

Find the \(\displaystyle x\)-bounds by finding the points of intersection of the two curves:

\[2x^2 = 1 + x^2 \implies x^2 = 1 \implies x = -1 \text{ or } x = 1\]

Set up the iterated integral:

\[\iint_D (x + 2y)\,dA = \int_{-1}^{1} \int_{2x^2}^{1+x^2} (x + 2y)\,dy\,dx\]

Evaluating the inner integral:

\[= \int_{-1}^{1} \left[xy + y^2\right]_{y=2x^2}^{y=1+x^2}\,dx\] \[= \int_{-1}^{1} \left[\left(x(1+x^2) + (1+x^2)^2\right) - \left(x(2x^2) + (2x^2)^2\right)\right]\,dx\]

This simplifies to:

\[= \int_{-1}^{1} \left[-3x^4 - x^3 + 2x^2 + x + 1\right]\,dx\]

Since \(\displaystyle -x^3\) and \(\displaystyle x\) are odd functions, they integrate to zero on \(\displaystyle [-1, 1]\). The remaining even function terms give:

\[= 2\int_0^1 \left(-3x^4 + 2x^2 + 1\right)\,dx = 2\left[-\frac{3x^5}{5} + \frac{2x^3}{3} + x\right]_0^1 = 2\left(-\frac{3}{5} + \frac{2}{3} + 1\right) = 2 \cdot \frac{16}{15} = \frac{32}{15}\]

Solution:

This region is best described as a Type II region. Fix \(\displaystyle y\): the left boundary is \(\displaystyle x = \frac{y^2 - 6}{2}\) (from the parabola) and the right boundary is \(\displaystyle x = y + 1\) (from the line).

Find the bounds for \(\displaystyle y\) by finding the points of intersection. Setting \(\displaystyle y + 1 = \frac{y^2 - 6}{2}\):

\[2(y + 1) = y^2 - 6 \implies y^2 - 2y - 8 = 0 \implies (y - 4)(y + 2) = 0\]

So \(\displaystyle y = -2\) and \(\displaystyle y = 4\).

Set up the iterated integral:

\[\iint_D xy\,dA = \int_{-2}^{4} \int_{\frac{y^2-6}{2}}^{y+1} xy\,dx\,dy\]

Evaluating the inner integral with respect to \(\displaystyle x\):

\[= \int_{-2}^{4} \left[\frac{x^2}{2} \cdot y\right]_{x=\frac{y^2-6}{2}}^{x=y+1}\,dy = \int_{-2}^{4} \left[\frac{(y+1)^2 y}{2} - \frac{(y^2-6)^2 y}{8}\right]\,dy\]

Solution:

First, sketch the region using the bounds given. From the original integral, fix \(\displaystyle x\): we have \(\displaystyle x \leq y \leq 1\) and \(\displaystyle 0 \leq x \leq 1\).

To switch the order, fix \(\displaystyle y\) instead. Reading from the diagram: \(\displaystyle 0 \leq x \leq y\) and \(\displaystyle 0 \leq y \leq 1\).

The switched integral becomes:

\[\int_0^1 \int_0^y \sin(y^2)\,dx\,dy\]

Evaluating the inner integral with respect to \(\displaystyle x\):

\[= \int_0^1 \left[x\sin(y^2)\right]_{x=0}^{x=y}\,dy = \int_0^1 y\sin(y^2)\,dy\]

Now use the substitution \(\displaystyle u = y^2\), \(\displaystyle du = 2y\,dy\):

\[= \left[-\frac{\cos(y^2)}{2}\right]_0^1 = -\frac{\cos(1)}{2} + \frac{1}{2} = \frac{1 - \cos(1)}{2}\]

Solution:

To write \(\displaystyle \iint_D f(x,y)\,dy\,dx\), we fix \(\displaystyle x\) and integrate with respect to \(\displaystyle y\) first. The region requires two separate integrals because the lower boundary changes at \(\displaystyle x = 1\).

Left part (\(\displaystyle e^{-3} \leq x \leq 1\)): The bottom curve is \(\displaystyle y = -\ln x\) and the top is \(\displaystyle y = 3\). Note that \(\displaystyle 3 = -\ln x\) gives \(\displaystyle x = e^{-3}\).

Right part (\(\displaystyle 1 \leq x \leq e^3\)): The bottom curve is \(\displaystyle y = \ln x\) and the top is \(\displaystyle y = 3\). Note that \(\displaystyle 3 = \ln x\) gives \(\displaystyle x = e^3\).

Therefore:

\[\iint_D f(x,y)\,dy\,dx = \int_{e^{-3}}^{1} \int_{-\ln x}^{3} f(x,y)\,dy\,dx + \int_{1}^{e^3} \int_{\ln x}^{3} f(x,y)\,dy\,dx\]

Solution:

Fix \(\displaystyle y\) and integrate with respect to \(\displaystyle x\) first. From the boundary curves, we solve for \(\displaystyle x\) in terms of \(\displaystyle y\):

From \(\displaystyle y = \ln x\): \(\displaystyle x = e^y\). From \(\displaystyle y = -\ln x\): \(\displaystyle x = e^{-y}\).

For a fixed \(\displaystyle y\) with \(\displaystyle 0 \leq y \leq 3\), the left boundary is \(\displaystyle x = e^{-y}\) and the right boundary is \(\displaystyle x = e^y\).

\[\iint_D f(x,y)\,dA = \int_0^3 \int_{e^{-y}}^{e^y} f(x,y)\,dx\,dy\]