In Section 15.1, we learned to compute the volume under a surface $z=f(x,y)$ over a rectangular domain $R$. But what if the domain $D$ is not a rectangle? What if it's a triangle, a disk, or the region enclosed by two parabolas?
The formal definition is still based on a Riemann sum. We can enclose our general region $D$ in a large rectangle $R$, and then define a new function $F(x,y)$ such that:
We then define the double integral over $D$ as $\iint_D f(x,y) \, dA = \iint_R F(x,y) \, dA$.
This definition of $F(x,y)$ is just a formal trick. Here's a simple way to think about it:
The integral $\iint_R F(x,y) \, dA$ is the total volume of dough left in the pan, which is exactly the volume of the cookie piece we wanted: $\iint_D f(x,y) \, dA$.
Explore the Analogy: This graph shows the rectangular pan $R$ (orange) and the cookie-cutter region $D$ (purple). Click the circle next to the top function $z = f(x,y)$ to see the solid volume we are trying to find.
This formal definition is clumsy for computation, but it's what justifies our new method. Fortunately, just as Fubini's Theorem let us use iterated integrals for rectangles, we can use them for general regions, but with a new twist: the bounds of integration can be functions.
A region $D$ is called Type I if it lies between the graphs of two continuous functions of $x$.
A domain $D$ is Type I if it can be described as:
$$ D = \{ (x,y) \mid a \le x \le b, \quad g_1(x) \le y \le g_2(x) \} $$To integrate over a Type I region, we slice vertically first:
$$ \iint_D f(x,y) \, dA = \int_a^b \left[ \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \right] dx $$The bounds of the outer integral must be constants. The bounds of the inner integral can be variables (functions of the outer variable).
Illegal: $\displaystyle \int_{g_1(x)}^{g_2(x)} \int_a^b f(x,y) \, dx \, dy$. The final answer must be a number, not a function of $x$!
Evaluate $\iint_D (x+2y) \, dA$, where $D$ is the region bounded by the parabolas $y=2x^2$ and $y=1+x^2$.
1. Sketch the region and find bounds:
First, find the intersection points by setting the functions equal:
$$ 2x^2 = 1+x^2 \implies x^2 = 1 \implies x = -1 \text{ and } x = 1 $$For $x$ in $[-1, 1]$, $1+x^2 \ge 2x^2$. Thus, $g_2(x) = 1+x^2$ (top curve) and $g_1(x) = 2x^2$ (bottom curve).
The region $D$ is Type I: $D = \{ (x,y) \mid -1 \le x \le 1, \quad 2x^2 \le y \le 1+x^2 \}$.
2. Set up and solve the integral:
$$ \iint_D (x+2y) \, dA = \int_{-1}^1 \int_{2x^2}^{1+x^2} (x+2y) \, dy \, dx $$Inner integral (w.r.t. $y$): Treat $x$ as a constant.
$$ \int_{2x^2}^{1+x^2} (x+2y) \, dy = \left[ xy + y^2 \right]_{y=2x^2}^{y=1+x^2} $$ $$ = \left( x(1+x^2) + (1+x^2)^2 \right) - \left( x(2x^2) + (2x^2)^2 \right) $$ $$ = (x+x^3 + 1+2x^2+x^4) - (2x^3 + 4x^4) $$ $$ = -3x^4 - x^3 + 2x^2 + x + 1 $$Outer integral (w.r.t. $x$):
$$ \int_{-1}^1 (-3x^4 - x^3 + 2x^2 + x + 1) \, dx = \left[ -\frac{3}{5}x^5 - \frac{1}{4}x^4 + \frac{2}{3}x^3 + \frac{1}{2}x^2 + x \right]_{-1}^1 $$This is $(-\frac{3}{5} - \frac{1}{4} + \frac{2}{3} + \frac{1}{2} + 1) - (\frac{3}{5} - \frac{1}{4} - \frac{2}{3} + \frac{1}{2} - 1) = -\frac{6}{5} + \frac{4}{3} + 2 = \frac{-18+20+30}{15} = \frac{32}{15}$.
Explore the Graph: Click off the red circle and the two orange circles to see the domain $D$ in the $xy$-plane. This is a Type I region, where we integrate from the lower curve $y=2x^2$ to the upper curve $y=1+x^2$. Click the orange circles back on and move the slider $a$ to see the slices parallel to the $y$-axis.
Find the volume of the solid under the surface $z=x^2+y^2$ and above the region $D$ in the xy-plane bounded by $y=\sqrt{x}$ and $y=x$.
1. Sketch the region and find bounds:
Intersections: $\sqrt{x} = x \implies x = x^2 \implies x^2-x=0 \implies x(x-1)=0$. So $x=0$ and $x=1$.
In the interval $[0,1]$, $\sqrt{x} \ge x$. So, $g_2(x) = \sqrt{x}$ (top) and $g_1(x) = x$ (bottom).
The region is $D = \{ (x,y) \mid 0 \le x \le 1, \quad x \le y \le \sqrt{x} \}$.
2. Set up and solve the integral:
$$ V = \int_0^1 \int_x^{\sqrt{x}} (x^2+y^2) \, dy \, dx $$Inner: $\int_x^{\sqrt{x}} (x^2+y^2) \, dy = \left[ x^2y + \frac{1}{3}y^3 \right]_{y=x}^{y=\sqrt{x}} = \left( x^2\sqrt{x} + \frac{1}{3}(\sqrt{x})^3 \right) - \left( x^2(x) + \frac{1}{3}x^3 \right) $
$$ = \left( x^{5/2} + \frac{1}{3}x^{3/2} \right) - \left( x^3 + \frac{1}{3}x^3 \right) = x^{5/2} + \frac{1}{3}x^{3/2} - \frac{4}{3}x^3 $$Outer: $\int_0^1 (x^{5/2} + \frac{1}{3}x^{3/2} - \frac{4}{3}x^3) \, dx = \left[ \frac{2}{7}x^{7/2} + \frac{1}{3}\cdot\frac{2}{5}x^{5/2} - \frac{4}{3}\cdot\frac{1}{4}x^4 \right]_0^1 $
$$ = \left[ \frac{2}{7} + \frac{2}{15} - \frac{1}{3} \right] - 0 = \frac{30 + 14 - 35}{105} = \frac{9}{105} = \frac{3}{35}. $$Explore the Graph: Click the circle next to the top function ($z=x^2+y^2$) to see the solid volume from Example 2.2.
Set up, but do not evaluate, the double integral $\iint_D (x\cos y) \, dA$ where $D$ is the region bounded by $y=0$, $y=x^2$, and $x=1$.
The region is bounded by $x=1$ on the right, $y=0$ on the bottom, and $y=x^2$ on the top. The intersection of $y=x^2$ and $y=0$ is at $x=0$.
This is a Type I region: $D = \{ (x,y) \mid 0 \le x \le 1, \quad 0 \le y \le x^2 \}$.
The integral is:
$$ \int_0^1 \int_0^{x^2} (x\cos y) \, dy \, dx $$A region $D$ is called Type II if it lies between the graphs of two continuous functions of $y$.
A domain $D$ is Type II if it can be described as:
$$ D = \{ (x,y) \mid c \le y \le d, \quad h_1(y) \le x \le h_2(y) \} $$To integrate over a Type II region, we slice horizontally first:
$$ \iint_D f(x,y) \, dA = \int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) \, dx \right] dy $$Set up the integral for the volume from Example 2.2 (under $z=x^2+y^2$ over $D$ bounded by $y=\sqrt{x}$ and $y=x$) as a Type II region.
1. Sketch the region and find bounds:
The region is bounded by $y=x$ and $y=\sqrt{x}$. We must express the boundaries as functions of $y$ ($x=h(y)$).
The intersection points are $(0,0)$ and $(1,1)$, so the $y$-values range from $c=0$ to $d=1$.
For $y$ in $[0,1]$, $y \ge y^2$. So the "right curve" is $h_2(y)=y$ and the "left curve" is $h_1(y)=y^2$.
The region is $D = \{ (x,y) \mid 0 \le y \le 1, \quad y^2 \le x \le y \}$.
2. Set up the integral:
$$ V = \int_0^1 \int_{y^2}^y (x^2+y^2) \, dx \, dy $$(Note: Evaluating this integral will also yield $3/35$.)
Evaluate $\iint_D xy \, dA$ where $D$ is the region bounded by the line $y=x-1$ and the parabola $y^2=2x+6$.
1. Sketch the region and find bounds:
This region is complicated as Type I (the bottom curve is defined piecewise). It is much simpler as Type II.
We solve for $x$ in terms of $y$:
Find intersections: $y+1 = \frac{1}{2}y^2-3 \implies 2y+2 = y^2-6 \implies y^2-2y-8 = 0 \implies (y-4)(y+2)=0$. So $y=-2$ and $y=4$.
The region is $D = \{ (x,y) \mid -2 \le y \le 4, \quad \frac{1}{2}y^2-3 \le x \le y+1 \}$.
2. Set up and solve the integral:
$$ \int_{-2}^4 \int_{\frac{1}{2}y^2-3}^{y+1} xy \, dx \, dy $$Inner: $\int_{\frac{1}{2}y^2-3}^{y+1} xy \, dx = y \left[ \frac{1}{2}x^2 \right]_{\frac{1}{2}y^2-3}^{y+1} = \frac{y}{2} \left[ (y+1)^2 - (\frac{1}{2}y^2-3)^2 \right]$
$$ = \frac{y}{2} \left[ (y^2+2y+1) - (\frac{1}{4}y^4 - 3y^2 + 9) \right] = \frac{y}{2} \left[ -\frac{1}{4}y^4 + 4y^2 + 2y - 8 \right] $$ $$ = -\frac{1}{8}y^5 + 2y^3 + y^2 - 4y $$Outer: $\int_{-2}^4 (-\frac{1}{8}y^5 + 2y^3 + y^2 - 4y) \, dy = \left[ -\frac{1}{48}y^6 + \frac{1}{2}y^4 + \frac{1}{3}y^3 - 2y^2 \right]_{-2}^4 $
$$ = \left( -\frac{4096}{48} + \frac{256}{2} + \frac{64}{3} - 32 \right) - \left( -\frac{64}{48} + \frac{16}{2} - \frac{8}{3} - 8 \right) $$ $$ = \left( -\frac{256}{3} + 128 + \frac{64}{3} - 32 \right) - \left( -\frac{4}{3} + 8 - \frac{8}{3} - 8 \right) $$ $$ = \left( -\frac{192}{3} + 96 \right) - \left( -\frac{12}{3} \right) = (-64 + 96) - (-4) = 32+4 = 36. $$Find the volume of the tetrahedron bounded by the planes $x+2y+z=2$, $x=2y$, $x=0$, and $z=0$.
1. Identify the top surface and the region $D$:
The solid sits on the $xy$-plane ($z=0$), and its top is the plane $x+2y+z=2$. Solving for $z$ gives the height function:
$$ z = 2 - x - 2y $$The region $D$ is the "shadow" of the solid on the $xy$-plane. Its three edges come from the three remaining planes:
The lines $y=\frac{x}{2}$ and $y=1-\frac{x}{2}$ meet when $\frac{x}{2} = 1 - \frac{x}{2} \implies x = 1$, at the point $(1, \frac{1}{2})$.
Using vertical slices, $D$ is Type I:
$$ D = \left\{ (x,y) \;\middle|\; 0 \le x \le 1, \quad \frac{x}{2} \le y \le 1 - \frac{x}{2} \right\} $$2. Set up and solve the integral:
$$ V = \int_0^1 \int_{x/2}^{1-x/2} (2 - x - 2y) \, dy \, dx $$Inner: $\displaystyle \int_{x/2}^{1-x/2} (2 - x - 2y) \, dy = \Big[ 2y - xy - y^2 \Big]_{y=x/2}^{y=1-x/2}$
$$ = \left( 2 - x - x + \frac{x^2}{2} - 1 + x - \frac{x^2}{4} \right) - \left( x - \frac{x^2}{2} - \frac{x^2}{4} \right) $$ $$ = \left( 1 - x + \frac{x^2}{4} \right) - \left( x - \frac{3x^2}{4} \right) = 1 - 2x + x^2 = (1-x)^2 $$Outer:
$$ V = \int_0^1 (1-x)^2 \, dx = \left[ -\frac{(1-x)^3}{3} \right]_0^1 = 0 - \left( -\frac{1}{3} \right) = \frac{1}{3}. $$(Note: The hardest part of a problem like this is not the integration. It is finding the region $D$ by projecting the solid down onto the $xy$-plane. Always sketch $D$ first.)
Explore the Graph: Click off the red, green, and blue circles to reveal the purple domain triangle $D$ from Example 3.3.
Set up the integral $\iint_D (x\cos y) \, dA$ from the Topic 2 CYU (where $D$ is bounded by $y=0$, $y=x^2$, and $x=1$) as a Type II integral.
The region is $D = \{ (x,y) \mid 0 \le x \le 1, \quad 0 \le y \le x^2 \}$.
To express as Type II, we solve for $x$: $y=x^2 \implies x=\sqrt{y}$ (since $x \ge 0$).
The $y$ values range from $y=0$ to $y=1$ (the value of $y$ when $x=1$).
The left curve is $x=h_1(y)=\sqrt{y}$. The right curve is $x=h_2(y)=1$.
The region is $D = \{ (x,y) \mid 0 \le y \le 1, \quad \sqrt{y} \le x \le 1 \}$.
The integral is:
$$ \int_0^1 \int_{\sqrt{y}}^1 (x\cos y) \, dx \, dy $$One of the most powerful applications of Type I and Type II regions is reversing the order of integration. Why would we do this?
Sometimes, an integral is difficult or impossible to compute in one order, but becomes simple in the other order. For example, $\int \sin(y^2) \, dy$ and $\int e^{x^2} \, dx$ do not have elementary antiderivatives.
Evaluate the integral $\displaystyle \int_0^1 \int_x^1 \sin(y^2) \, dy \, dx$.
1. Sketch the Region:
The integral is $dy \, dx$, so it's Type I. The bounds are:
This describes the triangle bounded by $y=x$ (bottom), $y=1$ (top), and $x=0$ (left). The vertices are (0,0), (0,1), and (1,1).
2. Re-describe as Type II:
We need $dx \, dy$ bounds (horizontal slices).
The region is $D = \{ (x,y) \mid 0 \le y \le 1, \quad 0 \le x \le y \}$.
3. Rewrite and Solve:
$$ \int_0^1 \int_0^y \sin(y^2) \, dx \, dy $$Inner: $\int_0^y \sin(y^2) \, dx = \sin(y^2) \left[ x \right]_0^y = y\sin(y^2)$.
Outer: $\int_0^1 y\sin(y^2) \, dy$. (Now it's solvable! Use u-sub: $u=y^2, du=2y\,dy$)
$$ = \frac{1}{2} \int_0^1 \sin(u) \, du = \frac{1}{2} \left[ -\cos(u) \right]_0^1 = -\frac{1}{2} (\cos(1) - \cos(0)) = \frac{1}{2}(1 - \cos(1)). $$Evaluate $\displaystyle \int_0^8 \int_{\sqrt[3]{y}}^2 e^{x^4} \, dx \, dy$.
1. Sketch the Region:
The integral is $dx \, dy$, so it's Type II. The bounds are:
The boundaries are $x=\sqrt[3]{y} \implies y=x^3$, $x=2$, and $y=0$. This is the region under the curve $y=x^3$ from $x=0$ to $x=2$. (Note: when $x=2$, $y=2^3=8$, which matches our $y$-bound).
2. Re-describe as Type I:
We need $dy \, dx$ bounds (vertical slices).
3. Rewrite and Solve:
$$ \int_0^2 \int_0^{x^3} e^{x^4} \, dy \, dx $$Inner: $\int_0^{x^3} e^{x^4} \, dy = e^{x^4} \left[ y \right]_0^{x^3} = x^3e^{x^4}$.
Outer: $\int_0^2 x^3e^{x^4} \, dx$. (Use u-sub: $u=x^4, du=4x^3\,dx$)
$$ = \frac{1}{4} \int_0^{16} e^u \, du = \frac{1}{4} \left[ e^u \right]_0^{16} = \frac{1}{4}(e^{16} - e^0) = \frac{1}{4}(e^{16} - 1). $$Reverse the order of integration for $\displaystyle \int_1^2 \int_0^{\ln x} f(x,y) \, dy \, dx$.
1. Sketch Region (Type I):
Bounds: $0 \le y \le \ln x$ and $1 \le x \le 2$.
Boundaries are $y=\ln x$, $y=0$, $x=1$ (note $\ln(1)=0$), and $x=2$.
2. Re-describe (Type II):
Solve $y=\ln x$ for $x \implies x=e^y$.
Left bound: $x=e^y$. Right bound: $x=2$.
Constant $y$ bounds: $y$ goes from $0$ to $\ln(2)$ (the value of $y$ when $x=2$).
3. Rewrite Integral:
$$ \int_0^{\ln 2} \int_{e^y}^2 f(x,y) \, dx \, dy $$The properties of double integrals (like linearity) still hold for general regions. We also gain two important applications.
This is extremely useful for regions that are neither Type I nor Type II, but can be split into Type I or Type II pieces.
Use a double integral to find the area of the region $D$ bounded by $y=x^2$ and $y=2x-x^2$.
Intersections: $x^2 = 2x-x^2 \implies 2x^2-2x=0 \implies 2x(x-1)=0$. So $x=0, x=1$.
In $[0,1]$, $2x-x^2 \ge x^2$. This is a Type I region.
We set up the integral for area, $f(x,y)=1$.
$$ \text{Area} = \int_0^1 \int_{x^2}^{2x-x^2} 1 \, dy \, dx $$Inner: $\int_{x^2}^{2x-x^2} 1 \, dy = [y]_{x^2}^{2x-x^2} = (2x-x^2) - (x^2) = 2x-2x^2$.
Outer: $\int_0^1 (2x-2x^2) \, dx = \left[ x^2 - \frac{2}{3}x^3 \right]_0^1 = (1 - \frac{2}{3}) - 0 = \frac{1}{3}$.
(Note: This is the exact same result as $\int_0^1 (f_{top} - f_{bottom}) \, dx$ from Calculus I. The double integral is the formal justification for that method.)
Set up the integral(s) for $\iint_D f(x,y) \, dA$ where $D$ is the triangle with vertices $(0,0)$, $(1,3)$, and $(5,1)$.
This region is neither Type I (the top boundary changes) nor Type II (the right boundary changes). We must split it.
Explore the Graph: Move the slider $a$ to sweep vertical slices across the triangle and watch the top boundary change at $x=1$ (Method 1). Then move the slider $b$ to sweep horizontal slices and watch the right boundary change at $y=1$ (Method 2).
Method 1: Split as Type I
We split the region at $x=1$.
Line 1 (from (0,0) to (1,3)): $y=3x$
Line 2 (from (0,0) to (5,1)): $y=x/5$
Line 3 (from (1,3) to (5,1)): $m = \frac{1-3}{5-1} = \frac{-2}{4} = -\frac{1}{2}$. So $y-1 = -\frac{1}{2}(x-5) \implies y = -\frac{1}{2}x + \frac{5}{2} + 1 \implies y = -\frac{1}{2}x + \frac{7}{2}$.
The integral is:
$$ \int_0^1 \int_{x/5}^{3x} f(x,y) \, dy \, dx + \int_1^5 \int_{x/5}^{-\frac{1}{2}x + \frac{7}{2}} f(x,y) \, dy \, dx $$Method 2: Split as Type II
Now slice horizontally. First, solve each line for $x$ in terms of $y$:
The left boundary is always Line 1, but the right boundary changes at the vertex $(5,1)$. So we split the region at $y=1$.
(Check: at $y=1$, both right curves give $x=5$, so the two pieces meet exactly at the vertex $(5,1)$.)
The integral is:
$$ \int_0^1 \int_{y/3}^{5y} f(x,y) \, dx \, dy + \int_1^3 \int_{y/3}^{7-2y} f(x,y) \, dx \, dy $$Sanity Check: With $f(x,y)=1$, both setups should give the area of the triangle. Method 1 gives $\frac{7}{5} + \frac{28}{5} = 7$, and Method 2 gives $\frac{7}{3} + \frac{14}{3} = 7$. ā
For this triangle, either order requires a split. Each split happens at the "middle" vertex: $x=1$ for vertical slices, $y=1$ for horizontal slices.
Set up a Type I double integral to find the area of the region bounded by $y=e^x$, $y=0$, $x=0$, and $x=1$.
The region is $D = \{ (x,y) \mid 0 \le x \le 1, \quad 0 \le y \le e^x \}$.
To find the area, we integrate $f(x,y)=1$.
$$ \text{Area} = \int_0^1 \int_0^{e^x} 1 \, dy \, dx $$(If we solve it: $\int_0^1 [y]_0^{e^x} \, dx = \int_0^1 e^x \, dx = [e^x]_0^1 = e^1 - e^0 = e-1$.)
Today we broke free from rectangular domains. We learned how to integrate over general regions by defining Type I (vertical) and Type II (horizontal) regions, which use functions as integration bounds.
The most important skills are sketching the domain $D$ and choosing the correct order of integration. Sometimes, an integral is impossible in one order but simple in the other, making reversing the order a critical problem-solving technique.
Find the volume of the solid under the plane $z=2x+y$ and above the region $D$ in the first quadrant bounded by $y=x$ and $y=x^3$.
1. Find Bounds:
Intersections: $x=x^3 \implies x^3-x=0 \implies x(x^2-1)=0$. In the first quadrant, $x=0$ and $x=1$.
In $[0,1]$, $x \ge x^3$. So $g_2(x)=x$ (top) and $g_1(x)=x^3$ (bottom).
2. Set up and Solve:
$$ V = \int_0^1 \int_{x^3}^x (2x+y) \, dy \, dx $$Inner: $\int_{x^3}^x (2x+y) \, dy = \left[ 2xy + \frac{1}{2}y^2 \right]_{y=x^3}^{y=x} $
$$ = \left( 2x(x) + \frac{1}{2}x^2 \right) - \left( 2x(x^3) + \frac{1}{2}(x^3)^2 \right) = \left( 2x^2 + \frac{1}{2}x^2 \right) - \left( 2x^4 + \frac{1}{2}x^6 \right) $$ $$ = \frac{5}{2}x^2 - 2x^4 - \frac{1}{2}x^6 $$Outer: $\int_0^1 (\frac{5}{2}x^2 - 2x^4 - \frac{1}{2}x^6) \, dx = \left[ \frac{5}{2}\cdot\frac{x^3}{3} - 2\frac{x^5}{5} - \frac{1}{2}\cdot\frac{x^7}{7} \right]_0^1 $
$$ = \left( \frac{5}{6} - \frac{2}{5} - \frac{1}{14} \right) - 0 = \frac{5 \cdot 35 - 2 \cdot 42 - 1 \cdot 15}{210} = \frac{175 - 84 - 15}{210} = \frac{76}{210} = \frac{38}{105}. $$Evaluate $\displaystyle\int_0^1 \int_{\sqrt{y}}^1 \sqrt{x^3+1} \, dx \, dy$.
The integral $\int \sqrt{x^3+1} \, dx$ is not elementary. We must reverse the order.
1. Sketch Region (Type II):
Bounds: $\sqrt{y} \le x \le 1$ and $0 \le y \le 1$.
Boundaries are $x=\sqrt{y} \implies y=x^2$, $x=1$, and $y=0$. Since $x \ge \sqrt{y}$ means $y \le x^2$, each horizontal slice runs from the parabola over to the line $x=1$. So $D$ is the region under the parabola $y=x^2$, above $y=0$, from $x=0$ to $x=1$.
2. Re-describe (Type I):
Vertical slices: $0 \le y \le x^2$ and $0 \le x \le 1$.
3. Rewrite and Solve:
$$ \int_0^1 \int_0^{x^2} \sqrt{x^3+1} \, dy \, dx $$Inner: $\int_0^{x^2} \sqrt{x^3+1} \, dy = \sqrt{x^3+1} [y]_0^{x^2} = x^2\sqrt{x^3+1}$.
Outer: $\int_0^1 x^2\sqrt{x^3+1} \, dx$. (Let $u=x^3+1, du=3x^2\,dx$)
$$ = \frac{1}{3} \int_1^2 \sqrt{u} \, du = \frac{1}{3} \left[ \frac{2}{3}u^{3/2} \right]_1^2 = \frac{2}{9} \left( 2^{3/2} - 1^{3/2} \right) = \frac{2}{9}(2\sqrt{2} - 1). $$Let $D$ be the triangular region with vertices $(0,0)$, $(1,2)$, and $(3,1)$.
1. Find the three edges:
Neither order avoids a split. The top boundary changes at the vertex $(1,2)$, and the right boundary changes at the vertex $(3,1)$.
(a) Type I: split at $x=1$.
(b) Type II: split at $y=1$.
(c) Evaluate (Type II is a little cleaner):
First piece: $\displaystyle \int_0^1 \int_{y/2}^{3y} y \, dx \, dy = \int_0^1 y\left(3y - \frac{y}{2}\right) dy = \int_0^1 \frac{5}{2}y^2 \, dy = \frac{5}{6}$.
Second piece:
$$ \int_1^2 \int_{y/2}^{5-2y} y \, dx \, dy = \int_1^2 y\left(5 - 2y - \frac{y}{2}\right) dy = \int_1^2 \left(5y - \frac{5}{2}y^2\right) dy $$ $$ = \left[ \frac{5}{2}y^2 - \frac{5}{6}y^3 \right]_1^2 = \left( 10 - \frac{20}{3} \right) - \left( \frac{5}{2} - \frac{5}{6} \right) = \frac{10}{3} - \frac{5}{3} = \frac{5}{3} $$Total: $\displaystyle \iint_D y \, dA = \frac{5}{6} + \frac{5}{3} = \frac{5}{6} + \frac{10}{6} = \frac{5}{2}$.
(The Type I setup gives the same answer, $\frac{35}{54} + \frac{100}{54} = \frac{5}{2}$.)