$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

# Double Integrals over General Regions

## Introduction

If the region over which we're evaluating a double integral is not rectangular, the order of integration matters, and the trickiest part of the problem tends to be finding the limits of integration.

There are two possibilities for this region $$R$$:

 Integrate with respect to $$y$$ first: Integrate with respect to $$x$$ first: $$\displaystyle\int_a^b \displaystyle\int_{g(x)}^{h(x)} f(x,y)\ dy\ dx$$ $$\displaystyle\int_c^d \displaystyle\int_{g(y)}^{h(y)} f(x,y)\ dx\ dy$$

Basically, we want to make sure that the last limits we plug in are constants, not functions.

## Examples

Evaluate the following integral.

$\int_0^1 \int_1^{x^2} 3x-y \ dy\ dx$

#### Solution

Note the order of integration: the inner integral has a limit in terms of x, and the outer integral is evaluated with respect to x.

\begin{align} \int_0^1 \int_1^{x^2} &3x-y \ dy\ dx\\ &= \int_0^1 \bigg[ 3xy-\dfrac{1}{2}y^2 \bigg]_1^{x^2}\ dx\\ &= \int_0^1 3x(x^2)-\dfrac{1}{2}(x^2)^2\ dx\\ &= \int_0^1 3x^3-\dfrac{1}{2}x^4\ dx\\ &= \dfrac{3}{4}x^4 - \dfrac{1}{10}x^5 \bigg|_0^1 = \ans{\dfrac{13}{20}} \end{align}

Evaluate $$\displaystyle\iint_R xy^2\ dA$$, where $$R$$ is the triangular region with vertices $$(0,0)$$, $$(2,4)$$, and $$(6,0)$$.

#### Solution

We could take vertical slices, but we'd have to split the region into two. Instead, we'll take horizontal slices (with $$\Delta y$$), so we'll integrate with respect to x first:

$\int A(y)\ dy = \iint f(x,y)\ dx\ dy$

The limits for y are from 0 to 4, and the limits for x are from the left-hand function to the right-hand one:

\begin{align} \iint_R xy^2\ dA &= \int_0^4 \int_{y/2}^{-y+6} xy^2\ dx\ dy\\ &= \int_0^4 \bigg[\dfrac{1}{2}x^2y^2\bigg]_{y/2}^{-y+6}\ dy\\ &= \int_0^4 \dfrac{3}{8}y^2-12y+36\ dy\\ &= \dfrac{1}{8}y^3-6y^2+36y \bigg|_0^4 = \ans{56} \end{align}

Evaluate $$\displaystyle\iint_R xy\ dA$$, where $$R$$ is bounded by $$y=2x+1$$, $$y=-2x+5$$, and $$x=0$$.

#### Solution

Here we'll take vertical slices, so we'll integrate with respect to y first:

\begin{align} \iint_R xy\ dA &= \int_0^1 \int_{2x+1}^{-2x+5} xy\ dy\ dx\\ &= \int_0^1 \bigg[\dfrac{1}{2}xy^2\bigg]_{2x+1}^{-2x+5}\ dx\\ &= \int_0^1 -12x^2+12x\ dx\\ &= -4x^3+6x^2 \bigg|_0^1 = \ans{2} \end{align}

Reverse the order of integration and evaluate $$\displaystyle\int_0^1 \displaystyle\int_{x^2}^1 x^3 \sin (y^3)\ dy\ dx$$.

#### Solution

\begin{align} \int_0^1 \int_0^{\sqrt{y}} x^3 \sin (y^3)\ dx\ dy &= \int_0^1 \dfrac{1}{4}y^2 \sin (y^3)\ dy\\ &= \ans{-\dfrac{1}{12}\cos 1 + 12} \end{align}

Find the volume of the solid under $$f(x,y)=x$$ bounded by $$y=x^2$$, $$y=-x+12$$, and $$y=4x+12$$.

#### Solution

Notice that the intersection points are $$-2$$ and $$3$$, but we need to split this into two regions, one from $$-2$$ to $$0$$, and the other from $$0$$ to $$3$$.

\begin{align} \textrm{Volume } &= \int_{-2}^0 \int_{x^2}^{4x+12} x\ dy\ dx + \int_0^3 \int_{x^2}^{-x+12} x\ dy\ dx\\ &= \int_{-2}^0 x(4x+12)-x(x^2)\ dx + \int_0^3 x(-x+12)-x(x^2)\ dx\\ &= \int_{-2}^0 -x^3+4x^2+12x\ dx + \int_0^3 -x^3-x^2+12x\ dx\\ &= \bigg[-\dfrac{1}{4}x^4+\dfrac{4}{3}x^3+6x^2 \bigg]_{-2}^0 + \bigg[-\dfrac{1}{4}x^4-\dfrac{1}{3}x^3+6x^2 \bigg]_0^3\\ &= \ans{\dfrac{185}{12}} \end{align}

#### Try it yourself:

(click on a problem to show/hide its answer)