Double integrals

What if you have an area that is bounded by limits on both the \(x\) and the \(y\)-axes? This is where double integrals come into play. We use double integrals in physics for finding mass in a region given a density function, engineering for determining properties of materials with varying thickness, and economics for analysing changes in multivariable contexts.

A double integral is of the form:

\[\underset{R}{\iint}f(x,y)dA=\underset{R}{\iint}f(x,y)dxdy=\underset{R}{\iint}f(x,y)dydx\]

where:

• \(z=f(x,y)\) is a continuous function on the region \(R\)
• \(R\) is a region of integration in the \(x\)-\(y\) plane
• \(dA\) is an infinitesimal element of area in \(R\) and \(dA=dxdy=dydx\).

If \(f(x,y)\geq0\), then the double integral is equal to the volume below the surface \(f(x,y)\) and above the region \(R\).

We can evaluate a double integral in two ways:

• Integrating first with respect to \(x\), keeping \(y\) constant, with \(a\leq x\leq b\) and \(c\leq y\leq d\):
  \[\underset{R}{\iint}f(x,y)dA=\int_{c}^{d}\left[\int_{a}^{b}f(x,y)dx\right]dy\]

  The "inner" integral is the integral with respect to \(x\) and the "outer" integral is the integral with respect to \(y\).

• Integrating first with respect to \(y\), keeping \(x\) constant, with \(a\leq y\leq b\) and \(c\leq x\leq d\):

  \[\underset{R}{\iint}f(x,y)dA=\int_{c}^{d}\left[\int_{a}^{b}f(x,y)dy\right]dx\]

  The "inner" integral is the integral with respect to \(y\) and the "outer" integral is the integral with respect to \(x\).

In practice, one may be easier than the other.

Evaluating double integrals

To evaluate a double integral, it is helpful to:

1. Sketch the region of integration.
2. Determine the limits of integration.
3. Evaluate the inner integral.
4. Evaluate the outer integral.

The most straightforward example is when the region \(R\) is a rectangle. This is covered in Example 1.

In many cases, the region of integration is not bounded by limits of integration that are constants; sometimes, they are functions. It is important to be able to sketch the region to identify the limits of integration. See Example 2 and 3 for how to do this.

Example 1 – evaluating double integrals

Evaluate \(\underset{R}{\iint}(3x^{2}-2xy)dxdy\), where \(R\) is the rectangle with vertices \((2,1),\) \((2,3),\) \((4,1),\) \((4,3)\).

First, sketch the region.

Let's look at both ways of integrating. In this example, you could reverse the order of integration and still obtain the same answer.

Integrating with respect to \(x\) then \(y\)

The "inner" integral is the integral with respect to \(x\) and the "outer" integral is the integral with respect to \(y\).

• The limits of integration for the inner integral are \(x=2\) to \(x=4\).
• The limits of integration for the outer integral are \(y=1\) to \(y=3\).

The integral becomes:
\[\int_{1}^{3}\left[\int_{2}^{4}(3x^{2}-2xy)dx\right]dy\]

First, we work out the inner integral, integrating with respect to \(x\) while treating \(y\) as a constant. Then, we integrate with respect to \(y\) and evaluate.
\[\begin{align*} \int_{1}^{3}\left[\int_{2}^{4}(3x^{2}-2xy)dx\right]dy & = \int_{1}^{3}\left[\left(\frac{3x^{3}}{3}-\frac{2x^{2}y}{2}\right)_{2}^{4}\right]dy\\
& = \int_{1}^{3}\left[\left(x^{3}-x^{2}y\right)_{2}^{4}\right]dy\\
& = \int_{1}^{3}\left[\left(4^{3}-4^{2}y\right)-\left(2^{3}-2^{2}y\right)\right]dy\\
& = \int_{1}^{3}\left[\left(64-16y\right)-\left(8-4y\right)\right]dy\\
& = \int_{1}^{3}\left[56-12y\right]dy\\
& = \left[56y-\frac{12y^{2}}{2}\right]_{1}^{3}\\
& = \left[56y-6y^{2}\right]_{1}^{3}\\
& = \left(56(3)-6(3)^{2}\right)-\left(56(1)-6(1)^{2}\right)\\
& = 168-54-56+6\\
& = 64
\end{align*}\]

Integrating with respect to \(y\) then \(x\)

In this case, the integral becomes:
\[\int_{2}^{4}\left[\int_{1}^{3}(3x^{2}-2xy)dy\right]dx\]

First, we work out the inner integral, integrating with respect to \(y\) while treating \(x\) as a constant. Then, we integrate with respect to \(x\) and evaluate.
\[\begin{align*} \int_{2}^{4}\left[\int_{1}^{3}(3x^{2}-2xy)dy\right]dx & = \int_{2}^{4}\left[\left(3x^{2}y-2x\frac{y^{2}}{2}\right)_{1}^{3}\right]dx\\
& = \int_{2}^{4}\left[\left(3x^{2}y-xy^{2}\right)_{1}^{3}\right]dx\\
& = \int_{2}^{4}\left[\left(3x^{2}(3)-x(3)^{2}\right)-\left(3x^{2}(1)-x(1)^{2}\right)\right]dx\\
& = \int_{2}^{4}\left[\left(9x^{2}-9x\right)-\left(3x^{2}-x\right)\right]dx\\
& = \int_{2}^{4}\left[6x^{2}-8x\right]dx\\
& = \left[\frac{6x^{3}}{3}-\frac{8x^{2}}{2}\right]_{2}^{4}\\
& = \left[2x^{3}-4x^{2}\right]_{2}^{4}\\
& = \left(2(4)^{3}-4(4)^{2}\right)-\left(2(2)^{3}-4(2)^{2}\right)\\
& = 128-64-16+16\\
& = 64
\end{align*}\]

As expected, the same answer is obtained regardless of the order of integration.

Example 2

Evaluate \(\underset{R}{\iint}(10x-4y+5)dxdy\), where \(R\) is the region bounded by the curve \(y=\sqrt{x}\) and the \(x\)-axis, from \(x=0\) to \(x=1\).

Integrating with respect to \(x\) then \(y\)

First, sketch the region of integration on the \(x-y\) plane. The "inner" integration is with respect to \(x\), so we draw an arrow across the region of integration, parallel to the \(x\)-axis (keeping \(y\) constant) from left to right.

Now, we can find the limits of integration. The left boundary is \(x=y^{2}\) (making \(x\) the subject of the equation \(y=\sqrt{x}\)). The right boundary is \(x=1\). These are the limits of integration for \(x\).

The outer integration is with respect to \(y\). The limits of integration are simply the lower and upper limits of \(y\) for the region of integration. The lower limit is \(y=0\) and the upper limit is \(y=1\).

Therefore, the integral becomes:
\[\int_{0}^{1}\left[\int_{x=y^{2}}^{1}(10x-4y+5)dx\right]dy\]

Now, we can integrate with respect to \(x\), and then with respect to \(y\).

\[\begin{align*} \int_{0}^{1}\left[\int_{x=y^{2}}^{1}(10x-4y+5)dx\right]dy & =\int_{0}^{1}\left[\left(5x^{2}-4yx+5x\right)_{x=y^{2}}^{1}\right]dy\\
& = \int_{0}^{1}\left[\left(5(1)^{2}-4y(1)+5(1)\right)-\left(5(y^{2})^{2}-4y(y^{2})+5(y^{2})\right)\right]dy\\
& = \int_{0}^{1}\left[\left(5-4y+5\right)-\left(5y^{4}-4y^{3}+5y^{2}\right)\right]dy\\
& = \int_{0}^{1}\left[-5y^{4}+4y^{3}-5y^{2}-4y+10\right]dy\\
& = \left[-y^{5}+y^{4}-\frac{5y^{3}}{3}-2y^{2}+10y\right]_{0}^{1}\\
& = \left(-(1)^{5}+(1)^{4}-\frac{5(1)^{3}}{3}-2(1)^{2}+10(1)\right)-\left(0\right)\\
& = -1+1-\frac{5}{3}-2+10\\
& = \frac{19}{3}\textrm{ or }6\frac{1}{3}
\end{align*}\]

Integrating with respect to \(y\) then \(x\)

Again, we can sketch the region of integration. The "inner" integration is with respect to \(y\), so we draw an arrow across the region of integration, parallel to the \(y\)-axis (keeping \(x\) constant) from bottom to top.

Now, we can find the limits of integration. The lower limit is \(y=0\) and the upper limit is \(y\sqrt{x}\). These are the limits of integration for \(y\).

The outer integration is with respect to \(x\). The limits of integration are simply the left and right limits of \(x\) for the region of integration. The left limit is \(x=0\) and the right limit is \(x=1\).

Therefore, the integral becomes:
\[\int_{0}^{1}\left[\int_{0}^{y=\sqrt{x}}(10x-4y+5)dy\right]dx\]

We integrate with respect to \(y\) and then with respect to \(x\). The answer is the same as integrating with respect to \(x\) first.

Example 3

Determine the limits of integration for\(\underset{R}{\iint}(2xy)dA\) over the region shown by:

1. integrating with respect to \(x\) first
2. integrating with respect to \(y\) first.

Integrating with respect to \(x\) then \(y\).

We draw an arrow parallel to the \(x\)-axis.

The left limit is \(x=-\dfrac{1}{3}(y-4)\), found by rearranging the equation \(y=-3x+4\)). The right limit is \(x=y\).

The \(y\) limits are the lower and upper values of \(y\). The lower limit is \(y=1\) and the upper limit is \(y=4\).

Therefore, the integral becomes:
\[\int_{1}^{4}\left[\int_{x=-\frac{1}{3}(y-4)}^{x=y}(2xy)dx\right]dy\]

Integrating with respect to \(y\) then \(x\).

In this case, we need to divide the region in two and draw two lines parallel to the \(y\)-axis.

Although the upper limit is always \(y=4\), the lower limit changes from \(y=-3x+4\) (between \(x=0\) and \(x=1)\), to \(y=x\) (from \(x=1\) to \(x=4\)).

This gives two double integrals that must be added together to cover the entire region.
\[\int_{0}^{1}\left[\int_{y=-3+4}^{y=4}(2xy)dy\right]\ dx+\int_{1}^{4}\left[\int_{y=x}^{y=4}(2xy)dy\right]dx\]

In this case, it appears that integrating with respect to \(x\) first is the easier option. In fact, sometimes one order may be impossible, but if the order of integration is reversed then the double integral can be evaluated.

Exercise – evaluating double integrals

1. Evaluate the following.
1. \(\underset{R}{\iint}(4xy)dxdy\), where \(R\) is the rectangle with vertices \((0,1)\), \((0,3)\), \((3,1)\) and \((3,3)\)
2. \(\underset{R}{\iint}(6e^{x}+e^{y})dydx\), where \(R\) is the region bounded by the lines \(x=0\), \(x=4\), \(y=1\) and \(y=2\)
3. \(\int_{0}^{3}\int_{1}^{4}(\sin(\pi x)+\cos(\pi y)dydx\).
2. Find \(\int_{0}^{b}\int_{0}^{a}(5)dxdy\). What is the physical significance of this result?
3. Evaluate the following double integrals.
1. \(\int_{0}^{1}\int_{y=x^{2}}^{y=x}(x^{3}y)dydx\)
2. \(\int_{0}^{2}\int_{y=x^{2}}^{y=4}(3+2y)dydx\)
3. \(\underset{R}{\iint}(4xy)dA\), where \(R\) is the region bounded by the graphs of \(y=x\) and \(y=x^{2}-2x\)
4. \(\underset{R}{\iint}(2x+y)dydx\), where \(R\) is the region bounded by the graphs of \(y+x=2\) and \(y=x^{2}\).
4. Consider the following:
   \[\int_{0}^{2}\left[\int_{y=0}^{y=x}(2x+2y)dy\right]dx\]
1. Evaluate the double integral.
2. Sketch the region of integration defined by the integral.
3. Confirm your answer by evaluating the double integral obtained by reversing the order of integration.
5. Consider the following:
   \[\int_{0}^{1}\left[\int_{y=x}^{y=1}\left(2x\frac{e^{y}}{y^{2}}\right)dy\right]dx\]
1. Sketch the region of integration defined by the integral.
2. Evaluate the double integral.

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