# Double Integrals – Examples with Answers

A double integral is a mathematical concept for finding the volume or area of a three-dimensional object or region in the plane. Double integrals can be expressed in rectangular or polar coordinates and can be evaluated using a variety of techniques.

In this article, we will give an overview of double integrals and show how to evaluate them using examples. In addition, we will look at some practice problems.

##### CALCULUS

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Learning about double integrals with examples.

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##### CALCULUS

Relevant for

Learning about double integrals with examples.

See examples

## How to solve double integrals?

To find a double integral, we must first identify a region in the plane over which we want to integrate. This region can be defined by inequalities or by drawing a limit curve.

Recall that the integral of a function represents the area under the curve. In the case of double integrals, the integral is the volume under a surface:

$$\int \int_{R} f(x, y)dA$$

Here, R is the region over which we integrate and $latex dA$ is the area element, an infinitesimal version of $latex \delta A$.

$latex dA$ depends on the coordinates. In Cartesian coordinates, we have:

Clearly, we see that $latex \delta A=\delta x \delta y$, then, we have $latex dA=dxdy$.

Therefore, this integral is evaluated as:

$$\int \int_{R} f(x, y)dA= \int \int_{R} f(x, y)dxdy$$

Now, suppose that the region of integration R is a rectangle defined by $latex a\leq x \leq b$ and $latex c\leq y \leq d$. Thus, the integral is:

$$\int \int_{R} f(x, y)dxdy= \int_{y=c}^{d} \left(\int_{x=a}^{b} f(x, y)dx \right)dy$$

First, we evaluate the inner integral and then the outer integral. In this case, the inner integral is with respect to $latex x$.

However, the order of integration doesn’t matter, since we could evaluate the integral with respect to $latex y$ first.

## Double integrals – Examples with answers

### EXAMPLE 1

In the following double integral, indicate the region $latex R$ of integration and find its value:

$$\int_{0}^{2} \int_{0}^{1}(1+2x+2y)dy dx$$

The integration region is a rectangle with vertices: $latex (0,0)$, $latex (2,0)$, $latex (0,1)$ and $latex (2,1)$.

Now, to solve the integral, we start with the innermost one, which in this case, is the one corresponding to the variable $latex y$:

$$\int_{0}^{2} \int_{0}^{1}(1+2x+2y)dy dx=\int_{0}^{2}\left[ \int_{0}^{1}(1+2x+2y)dy \right]dx$$

Therefore, the following simple integral is calculated with respect to $latex y$, treating the variable $latex x$ as if it were constant:

$$\int_{0}^{1}(1+2x+2y)dy=\int_{0}^{1}dy+2x\int_{0}^{1}dy+2\int_{0}^{1}ydy$$

$$=y\Big|_{0}^{1}+2xy\Big|_{0}^{1}+y^2\Big|_{0}^{1}$$

$$=(1-0)+2x(1-0)+(1^2-0^2)$$

$$=1+2x+1=2+2x$$

The result obtained is substituted between the brackets of the integral given at the beginning:

$$\int_{0}^{2}\left[ \int_{0}^{1}(1+2x+2y)dy \right]dx=\int_{0}^{2}(2+2x)dx$$

And the resulting integral is solved with respect to the variable $latex x$:

$$\int_{0}^{2}(2+2x)dx=2\int_{0}^{2}dx+2\int_{0}^{2}xdx$$

$$2x\Big|_{0}^{2}+x^2\Big|_{0}^{2}=2(2-0)+(2^2-0^2)$$

$$=4+4=8$$

### EXAMPLE 2

Calculate the double integral:

$$\iint(1+6xy^2)dxdy$$

On the flat region consisting of the rectangle given by: $latex R=[0,2] \times [-1,1]$.

The integral must be calculated over the limits given by $latex R$, choosing a certain order (Fubini’s theorem), since the function is continuous in the region of integration.

Since the given region is a rectangle, the variable $latex x$ goes from $latex 0$ to $latex 2$, while the variable $latex y$ goes from $latex -1$ to $latex 1$.

Starting to integrate the variable $latex y$, one has:

$$\int_{0}^{2}\left [ \int_{-1}^{1}(1+6xy^{2})dy \right ]dx$$

First, the bracketed integral is solved, where the integration variable is $latex y$, while $latex x$ is kept constant:

$$\int_{-1}^{1}(1+6xy^{2})dy =\int_{-1}^{1}dy+6x\int_{-1}^{1} y^{2}dy$$

$$=y\Big|_{-1}^{1}+6x\left(\dfrac{y^3}{3}\right)\Big|_{-1}^{1}$$

$$=[1-(-1)]+6x\left[\dfrac{1^3}{3}-\dfrac{(-1)^3}{3}\right]$$

$$=2+4x$$

This result is now substituted in the integral given at the beginning, and integrated with respect to the variable $latex x$:

$$\int_{0}^{2}\left [ \int_{-1}^{1}(1+6xy^{2})dy \right ]dx=\int_{0}^{2}(2+4x)dx$$

$$=\int_{0}^{2}2dx+\int_{0}^{2}4xdx$$

$$=2x\Big|_{0}^{2}+4\left(\dfrac{x^2}{2}\right)\Big|_{0}^{2}$$

$$=2(2-0)+4\left(\dfrac{2^2}{2}-\dfrac{0^2}{2}\right)$$

$$=4+8=12$$

### EXAMPLE 3

Solve the following double integral:

$$\int_0^3\int_{4x/3}^{\sqrt{25-x^2}}xdydx$$

The region of integration is the region between the straight line $latex y=\dfrac{4x}{3}$ and the circle $latex x^2+y^2=25$, as shown in the figure:

In said region, the intersection between the two curves is the point $latex P(3,4)$ and as the integral is established, it’s first integrated with respect to the variable $latex y$, which goes from the line $latex y = \dfrac{4x}{3}$ up to the circumference $latex +\sqrt{25-x^2}$, the variable $latex x$ remaining constant:

$$\int_0^3\int_{4x/3}^{\sqrt{25-x^2}}xdydx=\int_0^3x\left[\int_{4x/3}^{\sqrt{25-x^2}}dy\right]dx$$

The integral in square brackets is:

$$\int_{4x/3}^{\sqrt{25-x^2}}dy=\int_{4x/3}^{\sqrt{25-x^2}}dy=$$

$$=y\Big|_{4x/3}^{\sqrt{25-x^2}}=\sqrt{25-x^2}-\dfrac{4x}{3}$$

This result is then substituted into the original integral:

$$\int_0^3x\left[\int_{4x/3}^{\sqrt{25-x^2}}dy\right]dx=\int_0^3x\left[\sqrt{25-x^2}-\dfrac{4x}{3}\right]dx$$

$$=\int_0^3x\sqrt{25-x^2}dx-\int_0^3\dfrac{4x^2}{3}dx= I_1-I_2$$

Calculation of $latex I_1$.

The integral $latex I_1$ is solved by a change of variable:

• $latex u = 25-x^2$
• $latex du =-2xdx$

$$I_1 =\displaystyle\int_0^3x\sqrt{25-x^2}dx$$

$$=-\dfrac{1}{2}\displaystyle\int_{25}^{16} \sqrt{u}du$$

$$=\left(-\dfrac{1}{2}\right)\dfrac{u^\frac{3}{2}}{\frac{3}{2}}\Big|_{25}^{16}$$

$$=-\left(\dfrac{1}{3}\right){u^\frac{3}{2}}\Big|_{25}^{16}$$

$$=-\left(\dfrac{1}{3}\right)\left(\sqrt{16^3}-\sqrt{25^3}\right)=\dfrac{61}{3}$$

Calculation of $latex I_2$.

$$\displaystyle \int_0^3\dfrac{4x^2}{3}dx= \left(\dfrac{4}{3}\right)\dfrac {x^3}{3}\Big|_0^3=12$$

Therefore:

$$\int_0^3\int_{4x/3}^{\sqrt{25-x^2}}xdydx=\dfrac{61}{3}-12=\dfrac{25}{3}$$

### EXAMPLE 4

In the integral of example 3, change the order of integration and state the resulting integral.

In this case, the integration region would be divided into two parts, the first is a triangle whose vertices are $latex (0,0)$, $latex (0,4)$ and $latex (3,4)$, while the second is the sector above the black segment, the y-axis, and the circumference, as shown in the graph:

Therefore, an integral for each region has to be proposed, and then the respective results have to be added.

First, we integrated over the variable $latex x$, which goes from $latex x=0$ to the line $latex x =\dfrac{3y}{4}$, while the variable $latex y$ would go from $latex y=0$ to $latex y=4$.

Then, we have to add the region that goes from $latex x = 0$ to the circumference $latex x=\sqrt {25-y^2}$, in which $latex y$ varies from $latex y=4$ to $latex y=$5:

$$\int_0^3\int_{4x/3}^{\sqrt{25-x^2}}xdydx=\int_0^4\left[\int_{0}^{\frac{3y}{4}}xdx\right]dy+\int_4^5\left[\int_{0}^{\sqrt {25-y^2}}xdx\right]dy$$

### EXAMPLE 5

Solve the integrals in example 4 and compare with the result obtained in example 3.

$$\int_0^3\int_{4x/3}^{\sqrt{25-x^2}}xdydx=\int_0^4\left[\int_{0}^{\frac{3y}{4}}xdx\right]dy+\int_4^5\left[\int_{0}^{\sqrt {25-y^2}}xdx\right]dy=$$

$$=I_1+I_2$$

Calculation of $latex I_1$.

$$\displaystyle \int_0^4\left[\int_{0}^{\frac{3y}{4}}xdx\right]dy=\int_0^4\left[\dfrac{x^2}{2}\right]_0^{\frac{3y}{4}}dy$$

$$=\displaystyle\int_0^4\dfrac{9y^2}{32}dy=\left(\dfrac{3}{32}\right)y^3\Big|_0^4=6$$

Calculation of $latex I_2$.

$$\displaystyle\int_4^5\left[\int_{0}^{\sqrt {25-y^2}}xdx\right]dy=\int_4^5\dfrac{x^2}{2}\Big|_0^{\sqrt {25-y^2}}dy$$

$$=\dfrac{1}{2}\int_4^5 (25-y^2)dy=$$

$$=\dfrac{1}{2}\left[25y\Big|_4^5-\dfrac{y^3}{3}\Big|_4^5\right]$$

$$=\dfrac{1}{2}\left(25-\dfrac{61}{3}\right)=\dfrac{7}{3}$$

Therefore:

$$\int_0^3\int_{4x/3}^{\sqrt{25-x^2}}xdydx=6+\dfrac{7}{3}=\dfrac{25}{3}$$

This result is the same as that obtained in example 3, as expected.

### EXAMPLE 6

Solve the following:

$$\int\int_R xdA$$

Where $latex R$ is the region between $latex y=2x$ and $latex y=x^2$.

To solve the proposed integral, it’s necessary to identify $latex R$, which is shown in the figure on the left.

This is the shaded region between the straight line $latex y=2x$ and the parabola $latex y=x^2$.

One way to calculate the integral is to perform a vertical sweep. In this way, the function varies from the parabola $latex y=x^2$ to the straight line $latex y=2x$, while the variable $latex x$ varies from $latex x=0$ to $latex x=2$.

In that case, we have:

$$\int\int_R xdA=\int_0^2 x\left[\int_{x^2}^{2x}dy\right]dx$$

The integral is then solved in square brackets:

$$\int_{x^2}^{2x}dy=y\Big|_{x^2}^{2x}=2x-x^2$$

The result is then substituted and integrated with respect to $latex x$:

$$\int_0^2 x(2x-x^2)dx=\int_0^2 (2x^2-x^3)dx$$

$$=\int_0^2 2x^2dx-\int_0^2x^3dx$$

$$=\dfrac{2x^3}{3}\Big|_0^2-\dfrac{x^4}{4}\Big|_0^2$$

$$\left(\dfrac{2\cdot 2^3}{3}-\dfrac{2\cdot 0^3}{3}\right)-\left(\dfrac{2^4}{4}-\dfrac{0^4}{4}\right)$$

$$=\dfrac{16}{3}-4=\dfrac{4}{3}$$

The integral can also be calculated by making a horizontal sweep, in which case it is first integrated over the variable $latex y$, which varies between $latex \dfrac{y}{2}$ and $latex \sqrt {y}$, while $latex y$ varies between $latex y=0$ and $latex y=4$.

It is left as an exercise for the reader to verify that the result is the same.

### EXAMPLE 7

Solve the following:

$$\iint \limits_{R} x^{2}dx dy$$

Where R is the region between the x-axis and the parabola $latex 4-x^{2}$ in the interval $latex x\in [-2,2]$.

The integration region is shown in the figure above.

To solve the integral, a sweep is performed on the variable $latex y$, which goes from the x-axis to the parabola.

$$\displaystyle\iint \limits_{R} x^{2}dx dy=\int_{-2}^{2}x^{2}\left[ \int_{0}^{-x^{2}+4}dy\right] dx$$

The integral in square brackets results:

$$\displaystyle\int_{0}^{-x^{2}+4}dy = 4-x^{2}$$

It is then replaced in the original integral:

$$\displaystyle \int_{-2}^{2}x^{2}\left( 4-x^{2}\right) dx=\int_{-2}^{2}4x^{2}dx-\int_{-2}^{2}x^{4} dx$$

$$= \frac{4}{3} x^{3}\Big|_{-2}^{+2}-\frac{1}{5}x^{5} \Big|_{-2}^{+2}=\frac{128}{15}$$

### EXAMPLE 8

Find the volume of the solid in the first octant, on a rectangular base, bounded by:

i) The coordinate plane

ii) The plane x = 3

iii) The parabolic cylinder $latex z = 4 -y^2$

The volume of the solid is the integral under $latex z=f(x,y)$, with the walls given by the planes $latex y=0$, $latex z=0$ and $latex x=3$.

The first step is to determine the intersections of $latex f(x,y)=4-y^2$ with the coordinate axes, in order to find the limits of integration.

Intersection of $latex f(x,y)$ with z-axis

Taking $latex y = 0$, we have: $latex z =4-0^2=4$

Therefore, the intersection of the parabolic cylinder with the z-axis is the point $latex (0,0,4)$.

Intersection of $latex f(x,y)$ with y-axis

Taking $latex z= 0$, we have:

$latex 4-y^2=0\Rightarrow y =2$

The positive root is taken since the intersection with the positive y-axis is sought, which is the point $latex (0,2,0)$.

The plane $latex y=0$ limits the solid, since it is in the first octant, according to the statement, therefore, the base of the solid is a rectangle with dimensions $latex x=3$ and $latex y = 2$, while the two remaining walls are given by the planes $latex x=0$ and $latex x=3$.

The resulting solid is shown in the figure below:

The volume of the solid corresponds to the integral:

$$V=\int_a^b\int_c^d f(x,y)dydx$$

Where $latex y$ varies between $latex 0$ and $latex 2$, while $latex x$ varies between $latex 0$ and $latex 3$, therefore:

$$V=\int_0^3\left[\int_0^2 (4 -y^2)dy\right]dx$$

First, the inner integral is calculated:

$$\int_0^2 (4 -y^2)dy=\int_0^2 4dy-\int_0^2 y^2dy$$

$$4y\Big|_0^2-\dfrac{y^3}{3}\Big|_0^2$$

$$=8-\dfrac{8}{3}=\dfrac{16}{3}$$

The result is substituted in the volume integral bracket:

$$V=\int_0^3\left(\dfrac{16}{3}\right) dx$$

$$=\dfrac{16}{3}\int_0^3dx=\left(\dfrac{16}{3}\right)x\Big|_0^3=16$$

### EXAMPLE 9

Find the integral:

$$\displaystyle \iint \limits_{R} x \; y \; dx dy$$

Having the following:

$latex R = \{ (x,y) \in \mathbb{R}^{2} / \; 4-2x \leq y \leq 4-x^{2} \}$

The integral can be solved by performing a vertical sweep first, which means that the innermost integral is performed on the $latex y$ variable.

From the graph shown, the variable $latex y$ goes from the straight line $latex y =4-2x$ to the parabola $latex 4-x^{2}$. The intersections of both curves can be seen in the graph, they are the points $latex (0,4)$ and $latex (2,0)$, therefore, the variable $latex x$ goes from $latex x= 0$, to $latex x = 4$:

$$\displaystyle \iint \limits_{R} x \; y \; dx dy=$$

$$=\displaystyle\int_{0}^{2} \left[\int_{4-2x}^{4-x^{2}} x \; y \; dy \right]dx$$

$$=\int_{0}^{2} \left[ \; x \int_{4-2x}^{4-x^{2}} y dy \;\right] dx$$

The integral in square brackets is:

$$\left[ \; x \displaystyle\int_{4-2x}^{4-x^{2}} ydy \; \right]= \dfrac{1}{2}x^{2} ( x-2 ) ^{2} ( x+4 )$$

And substituting this result in the integral corresponding to the variable $latex x$, we obtain:

$$\displaystyle\int_{0}^{2}\left[\;\dfrac{1}{2}x^{2}( x-2) ^{2} ( x+4) \;\right]dx=\int_{0}^{2} \left( \dfrac{1}{2}x^{5}-6x^{3}+8x^{2} \right) dx$$

$$=\left[\dfrac{x^6}{12}-\dfrac{6x^4}{4}+\dfrac{8x^3}{3}\right]_0^2=\frac{8}{3}$$

### EXAMPLE 10

Calculate the volume of the solid bounded by the surfaces $latex z=4-x^2-2y^2$ and the plane $latex z=2$.

The volume to be calculated is below the paraboloid and above the $latex z=2$ plane, as shown in the figure above. It is given by the double integral:

$$V=\displaystyle\iint_R \left[f(x,y)-g(x,y)\right ]dxdy$$

Where R is the surface projected onto the $latex xy$ plane. In the example:

• $latex f(x,y) =4-x^2-2y^2$
• $latex g(x,y) =2$

Therefore:

$$V=\displaystyle\iint_R \left[(4-x^2-2y^2)-2\right]dxdy$$

Now we have to find the region of integration $latex R$, bounded by the curve resulting from the intersection of the surfaces. This is obtained by equating $latex f(x,y)=g(x,y)$:

$$4-x^2-2y^2=2$$

$$\Rightarrow -x^2-2y^2=-2= x^2+2y^2=2$$

Dividing by $latex 2$:

$$\dfrac{x^2}{2}+\dfrac{y^2}{1}=1$$

The intersection curve is an ellipse.

The intersections of said ellipse with the $latex x$ axis are $latex (-\sqrt{2},0$ and $latex (\sqrt{2},0$.

A sweep can be made over the y variable over the first quadrant so that the volume looks like this:

$$V=\displaystyle 4\int_ {0}^{\sqrt{2}}\int_ {0}^{\sqrt{1-\frac{x^2}{2}}}\left[(4-x^2-2y^2)-2\right]dxdy=$$

$$=\displaystyle 4\int_ {0}^{\sqrt{2}}\left[\int_ {0}^{\sqrt{1-\frac{x^2}{2}}}(2-x^2-2y^2) dy\right]dx$$

The integral in square brackets is:

$$\displaystyle \int_ {0}^{\sqrt{1-\frac{x^2}{2}}}(2-x^2-2y^2) dy=$$

$$= (2-x^2)\int_ {0}^{\sqrt{1-\frac{x^2}{2}}}dy-2\int_ {0}^{\sqrt{1-\frac{x^2}{2}}}y^2dy$$

$$=(2-x^2)\left(\sqrt{1-\frac{x^2}{2}}\right)-\dfrac{2}{3}\sqrt{\left(1-\frac{x^2}{2}\right)^3}$$

$$=\dfrac{(2-x^2)}{\sqrt 2}\sqrt{2-x^2}-\dfrac{2}{6\sqrt2}\sqrt{\left(2-x^2\right)^3}$$

$$=\dfrac{1}{\sqrt2}\sqrt{\left(2-x^2\right)^3}-\dfrac{2}{6\sqrt2}\sqrt{\left(2-x^2\right)^3}$$

$$=\left(\dfrac{1}{\sqrt2}-\dfrac{1}{3\sqrt2}\right)\sqrt{\left(2-x^2\right)^3}$$

$$=\dfrac{2}{3\sqrt2}\sqrt{\left(2-x^2\right)^3}$$

The result is substituted into the volume integral:

$$V=\displaystyle 4\int_ {0}^{\sqrt{2}}\dfrac{2}{3\sqrt2}\sqrt{\left(2-x^2\right)^3}dx$$

This integral is solved by trigonometric substitution:

• $latex x=\sqrt{2}\sin u$
• $latex dx=\sqrt{2}\cos u du$

The new limits of integration are calculated as follows:

For $latex {x =0}$

$latex \sqrt{2}\sin u=0\Rightarrow \sin u=0\Rightarrow u = 0$

For $latex {x =\sqrt{2}}$

$latex \sqrt{2}\sin u=\sqrt{2}\Rightarrow \sin u=1\Rightarrow u = \dfrac{\pi}{2}$

$$V=\displaystyle\dfrac{8}{3\sqrt2} \int_ {0}^{\frac{\pi}{2}}\sqrt{\left(2-2\sin^2u\right)^3}\sqrt{2}\cos u du=$$

$$=\displaystyle\dfrac{16\sqrt2}{3} \int_ {0}^{\frac{\pi}{2}}\sqrt{\left(1-\sin^2u\right)^3}\cos u du=\dfrac{16\sqrt2}{3} \int_ {0}^{\frac{\pi}{2}}\cos^4 u du$$

Using:

$$\displaystyle\int \cos^n x dx=\dfrac{\cos^{n-1}x\sin x}{n}+\dfrac{n-1}{n}\int\cos^{n-2}dx$$

We have:

$$\displaystyle\int \cos^4 x dx=\dfrac{\cos^{3}x\sin x}{4}+\dfrac{3}{4}\int\cos^{2}xdx$$

$$=\dfrac{\cos^{3}x\sin x}{4}+\dfrac{3}{8}(x+\cos x \sin x)+C$$

Therefore:
$$V=\dfrac{16\sqrt2}{3} \left[\dfrac{\cos^{3}u\sin u}{4}+\dfrac{3}{8}(u+\cos u \sin u)\right]_0^{\frac{\pi}{2}}$$

$$=\sqrt 2\pi$$

## Double integrals – Practice problems

Double integrals quiz  You have completed the quiz!

#### Find the integral of $latex f(x,y)=x^2-y^2$ over the rectangle with the limits $latex 1\leq x\leq 3$ and $latex 0\leq y\leq 2$.

Write the answer in the input box.

$latex I=$

Interested in learning more about integrals? You can take a look at these pages: ### Jefferson Huera Guzman

Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.  