In the following video, we will compute Example 1 and Example 2:

Example 1: Evaluate $$\iint_R f(x,y)\,dA,$$
where $f(x,y)= x^2y$ and $R$ is the upper half of the unit
disk (shown here).

Example 2: Evaluate $$\iint_R f(x,y)\,dA,$$
where $f(x,y) = 4xe^{2y}$ and the region $R$ is bounded by
the $x$-axis, the $y$-axis, the line $y=2$ and the curve
$y=\ln(x)$ (shown here).

Now, we look at some additional examples.

Example 3: Evaluate the integral $$I \iint _D\, (x+y)\,
dA$$ where $D$ consists of all points $(x,\,y)$ such that $0 \ \le \
y \ \le \sqrt{9-x^2}\,, \quad 0\ \le \ x \ \le 3\,.$

DO: Graph the region $D$ before reading
further.Then try to set up a Type I iterated
integral.

Solution 3: Since $y^2 \ = \ 9 -
x^2$ is a circle of radius $3$ centered at the origin, $D$
consists of all points in the first quadrant inside this
circle as shown here. This is described as a Type I
region, so we

fix $x$ in and integrate with respect to $y$ along
the black vertical line as shown, and

then integrate with respect to $x$ from $x=0$ to
$x=3$ (all possible black vertical lines).

So the double integral $I$ becomes the interated integral
\begin{eqnarray}
\int_0^3\left(\int_0^{\sqrt{9-x^2}}\, (x+y)\, dy\right)\,dx
&=& \int_0^3\, \Bigl[\, xy +\frac{1}{2}y^2
\,\Bigl]_0^{\sqrt{9 - x^2}}\,dx\\
&=&\int_0^3\, \Bigl( x\sqrt{9 - x^2} +\frac{1}{2} (9
- x^2)\Bigr)\, dx\\
\end{eqnarray}

DO: Evaluate the above
integral. The answer is 18.

Example 4: Evaluate the integral $\displaystyle
I=\iint_R\, (x+y)\, dA,$ where $R$ consists of all points $(x,y)$
with $0\le y\le 3$ and $0\le x\le\sqrt{9-y^2}$.

DO: Graph this region $R$.
Is it the same as $D$ above? Set up and evaluate the
integral above as a Type II integral over $R$, which will have
black horizontal lines. You should get the same answer as in
Example 3.