Fubini’s Theorem for Double Integrals

Direct Knowlege Contributor David A. Smith
  • By David A. Smith, Founder & CEO, Direct Knowledge
  • David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

Iterate Integrals Over Rectangular Regions

Green’s Theorem is named after the mathematician Guido Fubini.

Theorem. (Fubini’s Theorem for Rectangular Regions) If $f$ is a continuous function of $x$ and $y$ over the rectangle $R:$ $ a\leq x\leq b,$ $c\leq y\leq d,$ then the double integral of $f$ over $R$ may be evaluated by either iterated integral: \begin{align*} \iint_R f(x,y)d A & =\int_a^b \left[\int_c^d f(x,y) \, dy \right]dx \\ & =\int_c^d \left[\int_a^b f(x,y) \, dx\right]dy.\end{align*}

Fubini's Theorem for Double Integrals
The integral of $f(x,y)=x \sin xy$ over the region $R=[0,\pi]\times [0,1].$

Example. Evaluate the double integral \begin{equation} \iint_R x \sin x y \,dA \end{equation} over the region $R=\{ (x,y) \mid 0\leq x\leq \pi ,$ $0\leq y\leq 1\}.$

Solution. Since the integrand is continuous over $R$ we use Fubini’s Theorem to find \begin{align*} \iint_R x \sin x y \, dA & =\int_0^{\pi }\int_0^1 x \sin x y \, dy \, dx \\ & =-\int_0^{\pi } (\cos x-1) \, dx=\pi . \end{align*}

Fubini's Theorem for Double Integrals
The integral of $f(x,y)=(4+x^2)/(1+y^2)$ over the region $R=[0,1]\times [0,1].$

Example. Evaluate the double integral \begin{equation} \iint_R \left(\frac{4+x^2}{1+y^2}\right)d A \end{equation} over the region $R=\{ (x,y) \mid 0\leq x\leq 1 ,$ $0\leq y\leq 1\}.$

Solution. Since the integrand is continuous over $R$ we use Fubini’s Theorem to find \begin{align*} \iint_R \left(\frac{4+x^2}{1+y^2}\right)d A & = \int_0^1\int_0^1\left(\frac{4+x^2}{1+y^2}\right)dx dy \\ & =\int_0^1 \frac{13}{3 \left(y^2+1\right)} \, dy \\ & =\frac{13 \pi }{12}. \end{align*}

Fubini's Theorem for Double Integrals
The integral of $(2 x y)/(x^2+1)$ over the region $R=[0,1]\times [1,3].$

Example. Evaluate the double integral \begin{equation} \iint_R \frac{2 x y}{x^2+1}dA \end{equation} over the region $R=\{(x,y) \mid 0\leq x\leq 1, 1\leq y\leq 3\}.$

Solution. Since the integrand is continuous over $R$ we use Fubini’s Theorem to find \begin{align*} \iint_R \frac{2 x y}{x^2+1} \, dA & =\int_0^1\int_1^3\frac{2x y}{x^2+1} \, dy \, dx \\ & =\int_0^1 \frac{8 x}{x^2+1} \, dx \\ & =4 \ln (2). \end{align*}

Fubini's Theorem for Double Integrals
The integral of $f(x,y)=x^2e^{x y}$ over the region $R=[0,1]\times [0,1].$

Example. Evaluate the double integral \begin{equation} \iint_R \left(x^2e^{x y}\right)d A \end{equation} over the region $R=\{ (x,y) \mid 0\leq x\leq 1 ,$ $0\leq y\leq 1\}.$

Solution. Since the integrand is continuous over $R$ we use Fubini’s Theorem to find \begin{align*} \iint_R \left(x^2e^{x y}\right)d A & =\int_0^1\int_0^1x^2e ^{x y}dy dx \\ & = \int_0^1 \left(x e^x-x\right) \, dx \\ & =\frac{1}{2}. \end{align*}

Fubini's Theorem for Double Integrals
The integral of $f(x,y)=x/(x-y)$ over the region $R=[3,4]\times [1,2].$

Example. Evaluate the double integral \begin{equation} \int_3^4\int_1^2\frac{x}{x-y}dy\, dx. \end{equation}

Solution. Since the integrand is continuous over $R$ we use Fubini’s Theorem to find \begin{align*} \int_3^4\int_1^2\frac{x}{x-y} \, dy \, dx & =\int_3^4 x (\ln (1-x)-\ln (2-x)) \, dx \\ & =\frac{1}{2}-10 \ln (2)+\frac{15}{2} \ln (3). \end{align*}

Fubini's Theorem
The integral of $f(x,y)=x \sin xy$ over the region $R=[0,\pi]\times [0,1].$

Example. Evaluate the double integral \begin{equation} \int_2^3 \int_{-1}^2\frac{1}{(x+y)^2}dy\, dx. \end{equation}

Solution. Since the integrand is continuous over $R$ we use Fubini’s Theorem to find \begin{align*} \int_2^3\int_{-1}^2\frac{1}{(x+y)^2} \, dy \, dx & =\int_2^3 \left(\frac{1}{-x-2}-\frac{1}{1-x}\right) \, dx \\ & =\ln (2)+\ln (4)-\ln (5) \\ & =\ln \left(\frac{8}{5}\right). \end{align*} How does this example not contradiction Fubini’s Theorem?

Example. Show that the iterated integrals \begin{equation} \int_0^1 \int_0^1 \frac{y-x}{(x+y)^3} \, dy \, dx \quad \text{ and } \quad \int_0^1\int_0^1\frac{y-x}{(x+y)^3} \, dx \, dy \end{equation} have different values.

Solution. The first integral \begin{align*} \int_0^1\int_0^1\frac{y-x}{(x+y)^3} \, dy \, dx & =\int_0^1 -\frac{1}{(x+1)^2} \, dx \\ & =-\frac{1}{2}. \end{align*} and the second integral \begin{align*} \int_0^1\int_0^1\frac{y-x}{(x+y)^3} \, dx \, dy & =\int_0^1 \frac{1}{(y+1)^2} \, dy \\ & =\frac{1}{2}. \end{align*} do indeed have different values.

Iterated Integrals Over Non-Rectangular Regions

Definition. Let $R$ be a region in the $xy$-plane. Then $R$ is called a

(1) vertically simple region if $R$ can be described by the inequalities \begin{equation} D_1: \quad a\leq x\leq b, \quad g_1(x)\leq y\leq g_2(x) \end{equation} where $g_1(x)$ and $g_2(x)$ are continuous functions of $x$ on $[a,b].$

(2) horizontally simple region if $R$ can be described by the inequalities \begin{equation} D_2: \quad c\leq y\leq d, \quad h_1(y)\leq x\leq h_2(y) \end{equation} where $h_1(x)$ and $h_2(x)$ are continuous functions of $y$ on $[c,d].$ Vertically simple and horizontally simple regions:

vertically-simple-region
$D_1: \, a\leq x\leq b, \, g_1(x)\leq y\leq g_2(x)$
horizontally-simple-region
$D_2: \, c\leq y\leq d, \, h_1(y)\leq x\leq h_2(y)$

Theorem. (Fubini’s Theorem for Non-Rectangular Regions) If $D_1$ is a vertically simple region, then \begin{equation} \iint_{D_1} f(x,y) \, d A=\int_a^b \int_{g_1(x)}^{g_2(x)}f(x,y) \, dy \, dx \end{equation} whenever both integrals exist. Similarly, for a horizontally region $D_2,$ \begin{equation} \iint_{D_2} f(x,y) \, d A=\int_c^d \int_{h_1(y)}^{h_2(y)}f(x,y)\, dx \, dy\end{equation} whenever both integrals exist.

Example. Evaluate the double integral \begin{equation} \iint_D x y \, d A \end{equation} where $D$ is the region bounded by the line $y=x-1$ and the parabola $y^2=2x+6.$

Fubini's Theorem
Region of integration: $-2\leq y\leq 4$, $y^2/2-3\leq x\leq y+1.$

Solution. The region $D$ is both a vertically simple and a horizontally simple region, but the description of $D$ as a vertically simple region is more complicated because the lower boundary consists of two parts. Therefore we express $D$ as a horizontally simple region: \begin{equation} D=\left\{(x,y)\mid-2\leq y\leq 4,\frac{y^2}{2}-3\leq x\leq y+1\right\}. \end{equation} as shown. Then the double integral becomes \begin{align*} \iint_D x y \, d A & =\int_{-2}^4\int_{(1/2)y^2-3}^{y+1}x y \, dx \,dy \\ & =\frac{1}{2}\int_{-2}^4 \left(-\frac{y^5}{4}+4 y^3+2 y^2-8y\right) \, dy \\ & = 36. \end{align*} If we had expressed $D$ as a vertically simple region, then we would have obtained \begin{align*} \iint_D x y \, d A & =\int_{-3}^{-1}\int_{-\sqrt{2x+6}}^{\sqrt{2x+6}}x y \, dy \, dx+\int_{-1}^5\int_{x-1}^{\sqrt{2x+6}}x y \, dy \, dx \\ & =36. \end{align*}

Example. Evaluate the iterated integral $$ \int_0^1\int_{x^2}^{\sqrt{x}}x y^2 \, dy \, dx. $$

Fubini's Theorem
Region of integration: $0\leq x\leq 1$, $x^2 \leq y \leq \sqrt{x}.$

Solution. We express the region of integration $D$ as a vertically simple region: \begin{equation} D=\left\{(x,y)\mid 0\leq x\leq 1, x^2 \leq y \leq \sqrt{x} \right\}. \end{equation} We use Fubini’s Theorem to obtain \begin{align*} \int_0^1\int_{x^2}^{\sqrt{x}}x y^2 \, dy \, dx & =\int_0^1x \left(\frac{x^{3/2}}{3}-\frac{x^6}{3}\right) \, dx \\ & =\frac{3}{56}. \end{align*}

Fubini's Theorem
Region of integration: $0\leq y\leq 2\sqrt{3}$, $\frac{y^2}{6} \leq x \leq \sqrt{16-y^2}.$

Example. Evaluate the iterated integral $$ \int_0^{2\sqrt{3}} \int_{\left.y^2\right/6}^{\sqrt{16-y^2}} \, dx \, dy. $$

Solution. We express the region of integration $D$ as a horizontally simple region: \begin{equation} D=\left\{(x,y)\mid 0\leq y\leq 2\sqrt{3}, \frac{y^2}{6} \leq x \leq \sqrt{16-y^2} \right\}. \end{equation} We use Fubini’s Theorem to obtain \begin{align*}\int_0^{2\sqrt{3}} \int_{\left.y^2\right/6}^{\sqrt{16-y^2}} \, dx \, dy & =\int_0^{2\sqrt{3}}\left(\sqrt{16-y^2}-\frac{y^2}{6}\right) \, dy \\ & =\left(\frac{1}{18} \left(12 \sqrt{3}+48 \pi \right) \right). \end{align*}

Fubini's Theorem
Region of integration: $0\leq x\leq 1$, $-x^2 \leq y \leq x^2.$

Example. Evaluate the iterated integral $$ \int_0^1\int_{-x^2}^{x^2} \, dy \, dx. $$

Solution. We express the region of integration $D$ as a vertically simple region: \begin{equation} D=\left\{(x,y)\mid 0\leq x\leq 1, -x^2 \leq y \leq x^2 \right\}. \end{equation} The region of integration is sketched. We find \begin{align} & \int_0^1\int_{-x^2}^{x^2} \, dy \, dx =\int_0^12 x^2 \, dx =\frac{2}{3}. \end{align}

Fubini's Theorem
Region of integration: $-2\leq y\leq 1$ and $y^2+4y\leq x\leq 3y+2.$

Example. Sketch the region of integration and evaluate \begin{equation}
\int_{-2}^1\int_{y^2+4y}^{3y+2} \, dx \, dy. \end{equation}

Solution. The region is horizontally simple region as $-2\leq y\leq 1$ and $y^2+4y\leq x\leq 3y+2.$ We find \begin{align*} \int_{-2}^1\int{y^2+4y}^{3y+2} dx \, dy & =\int_{-2}^1 \left(-y^2-y+2\right) \, dy \\ & = \frac{9}{2}. \end{align*} This region could also be considered as a vertically simple region as $-4\leq x\leq 5$ and $\frac{x-2}{3}\leq y\leq -2+\sqrt{x+4}.$ We find \begin{align*} \int_{-4}^5\int{(x – 2) / 3}^{-2+\sqrt{x+4}} dy \, dx & =\int_{-4}^5 \left(\left(-2+\sqrt{x+4}\right)-\left(\frac{x-2}{3}\right)\right) \, dx \\ & =\frac{9}{2}. \end{align*}

Fubini's Theorem
Region of integration: $0\leq x\leq \frac{\pi }{2}$ and $0\leq y\leq \sin x.$

Example. Sketch the region of integration and evaluate \begin{equation} \int_0^{\pi /2}\int_0^{\sin x}e^y \cos x \, dy \, dx. \end{equation}

Solution. The region is vertically simple region as $0\leq x\leq \frac{\pi }{2}$ and $0\leq y\leq \sin x.$ We find \begin{align*} \int_0^{\pi /2}\int_0^{\sin x}e^y \cos x \, dy \, dx & =\int_0^{\pi /2} \left(e^{\sin x}\cos x-\cos x\right) \, dx \\ & =e-2.\end{align*} The region is also horizontally simple region as $0\leq y\leq 1$ and $\text{Arcsin}(y)\leq x\leq \frac{\pi }{2}.$ We find \begin{align*} \int_0^1\int_{\arcsin y }^{\pi /2}e^y \cos x \, dx \, dy =e-2. \end{align*}

Fubini's Theorem
Region of integration is: $0\leq x\leq 1$, $x\leq y\leq 1$ but also as $0\leq y\leq 1$, $0\leq x\leq y.$

Example. Sketch the region of integration and evaluate the iterated integral \begin{equation} \int_0^1\int_x^1\sin \left(y^2\right) \, dy \, dx. \end{equation}

Solution. If we try to evaluate the integral as it stands, we are faced with the task of first evaluating \begin{equation} \int \sin y^2 \,dy. \end{equation} We elect to change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. We have \begin{equation}
\int_0^1\int_x^1\sin y^2 \, dy \, dx=\iint_D \sin y^2 \, d A \end{equation} where $D=\{(x,y) \mid 0\leq x\leq 1,x\leq y\leq 1\}.$ This region has an alternate description: \begin{equation} D=\{(x,y) \mid 0\leq y\leq 1,0\leq x\leq y\}. \end{equation} Thus we can express the double integral as an iterated integral in the reverse order: \begin{align*} \int_0^1\int_x^1\sin y^2 \, dy \, dx & = \iint_D \sin y^2 \, d A \\ & =\int_0^1\int_0^y\sin y^2 \, dx \, dy \\ & =\int_0^1\left.\left[x \sin y^2\right]\right|{0}^{y}dy \\ & =\frac{1}{2}(1-\cos 1). \end{align*}

Volume as a Double Integral

Theorem. (Volume as a Double Integral) If $f(x,y)\geq 0$ on the rectangular region $R,$ then the product $f\left(x_k^*, y_k^* \right) \triangle A_k$ is the volume of a parallelepiped (a box) with height $f\left(x_k^*,y_k^*\right)$ and base area $\triangle A_k.$ The Riemann sum \begin{equation} \sum_{k=1}^N f\left(x_k^*, y_k^*\right) \triangle A_k \end{equation} provides an estimate of the total volume under the surface $z=f(x,y)$ over $R,$ and if $f$ is continuous, we expect the approximation to improve by using more refined partitions. That is, the volume under $z=f(x,y)$ over the domain $R$ is given by \begin{equation} \lim_{|P|\to 0}\sum_{k=1}^N f\left(x_k^*, y_k^*\right)\triangle A_k=\iint_R f(x,y)d A \end{equation} when $f(x,y)\geq 0$ on the rectangular region $R.$

Example. Find the volume of the bounded solid that lies inside both the cylinder $x^2+y^2=3$ and the sphere $x^2+y^2+z^2=7.$

Solution. The volume is given by \begin{equation} V=4\int_0^{\sqrt{3}}\int_0^{\sqrt{3-y^2}}\sqrt{7-x^2-y^2} \, dx \, dy\end{equation} or also \begin{equation} V=4\int_0^{\sqrt{3}}\int_0^{\sqrt{3-x^2}}\sqrt{7-x^2-y^2} \, dy \, dx\approx 22.0336. \end{equation}

Example. Find the volume of the solid region bounded below by the given rectangle in the $x y$-plane and above by the graph of the given surface \begin{equation}
f(x,y)=\frac{x y}{\sqrt{x^2+y^2+1}} \end{equation} on $0\leq x\leq 1, 0\leq y\leq 1$

Solution. The volume is given by \begin{align} \int_0^1\int_0^1\frac{x y}{\sqrt{x^2+y^2+1}} \, dy \, dx & =\int_0^1 x \left(\sqrt{x^2+2}-\sqrt{x^2+1}\right) \, dx \\ &=\frac{1-2 \sqrt{2}}{3}+\frac{-2 \sqrt{2}+3 \sqrt{3}}{3} \\ & =\frac{1-4 \sqrt{2}+3 \sqrt{3}}{3} \\ & \approx 0.179766 \end{align}

Example. Find the volume of the solid region bounded below by the given rectangle in the $x y$-plane and above by the graph of the surface $$ f(x,y)=(x+y)^5 $$ on $0\leq x\leq 1, 0\leq y\leq 1.$

Solution. The volume is \begin{align*} \int_0^1\int_0^1(x+y)^5dy\, dx =\int_0^1 \left(\frac{1}{6} (x+1)^6-\frac{x^6}{6}\right) \, dx =3. \end{align*}

Example. Find the volume of the bounded solid between the two elliptic paraboloids $$ z=\left.x^2\right/(9+y^2-4) \quad \text{and}\quad z=\left.-x^2\right/(9-y^2+4). $$

Solution. Using symmetry, the volume is given by \begin{align*} V& =8\int_0^6\int_0^{\sqrt{4-\left.x^2\right/9}}\int_0^{4-\left.x^2\right/9-y^2} \, dz \, dy \, dx \\ & =8\int_0^6\int_0^{\sqrt{4-\left.x^2\right/9}}\left(-\frac{x^2}{9}-y^2+4\right) \, dy \, dx \\ & =8\int_0^6\left(\frac{8}{3} \sqrt{4-\frac{x^2}{9}}-\frac{2}{27} x^2 \sqrt{4-\frac{x^2}{9}}\right) \, dx \\ &= 48 \approx 150.796. \end{align*}

Example. Find the volume of the bounded solid bounded below by the rectangle $ R: 1\leq x\leq 2, 1\leq y\leq 2$ in the $x y$-plane and above by the graph of \begin{equation} z=f(x,y)=\frac{x}{y}+\frac{y}{x}. \end{equation}

Solution. The volume is given by the following double integral and can be compute using Fubini’s theorem. \begin{align} \iint_R \left(\frac{x}{y}+\frac{y}{x}\right) \, dA & =\int_0^2\int_1^2\left(\frac{x}{y}+\frac{y}{x}\right) \, dy \, dx \\ & =\int_0^2 \left.\left[x \ln |y|+\frac{y^2}{2x}\right]\, \right|^2_1 dx \\ & =\left.\left[\frac{3}{2}\ln |x|\frac{+1}{2}(\ln 2)x^2\right]\, \right|^2_1 \\ &= \text{3 ln 2}. \end{align}

Exercises on Fubini’s Theorem

Exercise. Sketch the region of integration and evaluate $\int_{-2}^1\int_{y^2+4y}^{3y+2} \, dx \, dy.$

Exercise. Evaluate $\iint_R (x^2+y^2)^{3/2} \, dA$ where $R$ is the unit circle centered at $(0,0).$

Exercise. Evaluate the following iterated integrals.

$(1) \quad \displaystyle \int_0^2\int_0^1\left(x^2+x y+y^2\right)dy\, dx.$

$(2) \quad \displaystyle \int_1^2\int_0^{\pi }x \cos y \, dy\, dx.$

$(3) \quad \displaystyle \int_0^{\ln (2)}\int_0^1e^{x+2y}dx\, dy.$

$(4) \quad \displaystyle \int_3^4\int_1^2\frac{x}{x-y}dy\, dx$

$(5) \quad \displaystyle \int_2^3\int_{-1}^2\frac{1}{(x+y)^2}dy\, dx$

Exercise. Use iterated integration to compute the double integral of the given rectangular region for each of the following.

$(1) \quad R=\{(x,y)\mid1\leq x\leq 2,0\leq y\leq 1\}$ $$ \iint_R x^2y\, dA $$

$(2) \quad R=\{(x,y)\mid-1\leq x\leq 0,0\leq y\leq \ln (2)\}$ $$ \iint_R2x e^y\, dA $$

$(3) \quad R=\left\{(x,y)\mid 0\leq x\leq \frac{\pi }{4}, \, 0\leq y\leq \frac{\pi }{2}\right\}$ $$ \iint_R\sin (x+y)\, dA $$

$(4) \quad R=\{(x,y)\mid0\leq x\leq \pi ,0\leq y\leq 1\}$ $$ \iint_Rx \sin (x y)\, dA $$

$(5) \quad R=\{(x,y)\mid0\leq x\leq 1,0\leq y\leq 2\}$ $$ \iint_Rx\sqrt{1-x^2}e^{3y}\, dA $$

$(6) \quad R=\{(x,y)\mid0\leq x\leq 1,1\leq y\leq 2\}$ $$ \iint_Rx e^{x y}\, dA $$

Exercise. Find the volume of the solid bounded below by the given rectangular region in the $x y$-plane and above the graph of the given function.

$(1) \quad z=2x+3y$, $ R=\{(x,y)\mid0\leq x\leq 1,0\leq y\leq 2\} $

$(2) \quad z=x \ln ( x y)$, $ R=\{(x,y)\mid1\leq x\leq 2,1\leq y\leq e\} $

$(3) \quad z=x \cos y+y \sin x.$ $$ R=\left\{(x,y)\mid0\leq x\leq \frac{\pi }{2}\, ,0\leq y\leq \frac{\pi}{2}\right\} $$

Exercise. Find the volume of the solid region bounded below by the given rectangle in the $x y$-plane and above by the graph of the given surface.

$(1) \quad \displaystyle f(x,y)=\frac{x y}{\sqrt{x^2+y^2+1}}$ on $0\leq x\leq 1, 0\leq y\leq 1$

$(2) \quad f(x,y)=(x+y)^5$ on $0\leq x\leq 1, 0\leq y\leq 1$

Exercise. Sketch the region and evaluate the iterated integral over the non-rectangular region.

$(1) \quad \displaystyle \int_0^4\int_0^{4-x}x ydy\, dx.$

$(2) \quad \displaystyle \int_1^e\int_0^{\ln (x)}x ydy\, dx.$

$(3) \quad \displaystyle \int_0^2\int_0^{\sin (x)}y \cos xdy\, dx.$

$(4) \quad \displaystyle \int_{-2}^1\int_{y^2+4y}^{3y+2} dx\, dy.$

$(5) \quad \displaystyle \int_0^{2\sqrt{3}}\int_{\left.y^2\right/6}^{\sqrt{16-y^2}}
dx\, dy.$

$(6) \quad \displaystyle \int_0^1\int_{-x^2}^{x^2} dy\, dx.$

$(7) \quad \displaystyle \int_0^4\int_{x^2}^{4x} dy\, dx.$

$(8) \quad \displaystyle \int_0^1\int_{x^2}^{\sqrt{x}}x y^2dy\, dx$

$(9) \quad \displaystyle \int_{-1}^3\int_{\tan ^{-1}x}^{\pi /4}x ydy\, dx.$

$(10) \quad \displaystyle \int_0^4\int_0^{4-x}x ydy\, dx.$

$(11) \quad \displaystyle \int_0^7\int_{x^2-6x}^x \sin x dy\, dx$

Exercise. Find the volume of the solid bounded by $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1. $$ (Setup but do not evaluate the double integral)

Exercise. Find the volume of the solid that lies inside both the cylinder $x^2+y^2=3$ and the sphere $x^2+y^2+z^2=7.$ (Setup but do not evaluate the double integral).

Exercise. For each of the following sketch the region of integration and write an equivalent integral with the order of integration reversed for

$(1) \quad \displaystyle \int_0^1\int_0^{2y}f(x,y)\, dx\, dy.$

$(2) \quad \displaystyle \int_0^1\int_{x^2}^{\sqrt{x}}f(x,y)\, dy\, dx.$

$(3) \quad \displaystyle \int_0^4\int_{y/2}^{\sqrt{y}}f(x,y)\, dx\, dy.$

$(4) \quad \displaystyle \int_1^2\int_{\ln (x)}^2f(x,y)\, dy\, dx.$

$(5) \quad \displaystyle \int_0^3\int_{y/3}^{\sqrt{4-y}}f(x,y)\, dx\, dy.$

Exercise. Evaluate the double integral $$ \iint_D x y \, dA $$ where $D$ is the triangular region in the $x y$-plane with vertices $(0,0),$ $(1,0),$ and $(4,1).$

Exercise. Evaluate the double integral $$ \iint_D \left(x^2-x y-1\right)\, dA $$ where $D$ is the triangular region in the $x y$-plane bounded by the lines $x-2y+2=0,$ $x+3y-3=0,$ and $y=0.$

Exercise. (a) Write the iterated integral $ I=\int_0^1 \int_y^1 \sin ( x^2 ) \, dx \, dy$ as a double integral over a domain $R.$ (b) Sketch the domain $R.$ (c) Evaluate the double integral by reversing the order of integration.