Double Integral Calculator is a free online tool that displays the value of the double integral function. STUDYQUERIES’S online double integral calculator tool makes the calculation faster, and it displays the double integral value in a fraction of seconds.

What is Double Integral Calculator?

A double Integral Calculator is an online tool that helps to integrate a given function and obtain the value of the double integral. Double integrals can be used to find the volume under a surface and the average value of a function with two variables. To use the Double Integral Calculator, enter the values in the input boxes.

How to Use Double Integral Calculator?: Step By Step Guide

Using the online double integral calculator, please follow the steps below to find the value of the double integral:

  • Step 1: Access STUDYQUERIES’s online Double Integral Calculator.
  • Step 2: Enter the function and the limits into the given input boxes. Select the variable to be integrated first from the drop-down list.
  • Step 3: Press the “Calculate” button to find the value of the double integral.
  • Step 4: Click the “Reset” button to clear the fields and enter new information.

Double Integral Calculator

What Is Double Integral?

The definite integral can be extended to functions of more than one variable. Consider, for example, a function of two variables $$z=f(x,y)$$ The double integral of a function f(x,y) is denoted by

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

where R is the region of integration in the xy-plane.

If the definite integral $$\int_{a}^{b}f(x)dx$$

of a function of one variable $$f(x)\geq 0$$ is the area under the curve f(x) from x = a to x = b then the double integral is equal to the volume under the surface z = f(x,y) and above the xy-plane in the region of integration R.

Double Integral Calculator
Double Integral Calculator

As in the case of the integral of a function of one variable, a double integral is defined as a limit of a Riemann sum.

If the region is $$R$$ a rectangle $$[a,b]\times [c,d]$$, we can subdivide $$[a,b]$$ into small intervals with a set of numbers $${x_0,x_1,\cdots x_m}$$ so that $$a= x_0\lt x_1\lt x_2\lt \cdots \lt x_i\lt \cdots \lt x_{m-1}\lt x_m=b$$

Partition Of A Rectangular Region Of Integration Into Small Intervals
Partition Of A Rectangular Region Of Integration Into Small Intervals

Similarly, a set of numbers $${y_0,y_1,\cdots y_m}$$ is said to be a partition of [c,d] along the y-axis, if

$$c= y_0\lt y_1\lt y_2\lt \cdots \lt y_i\lt \cdots \lt y_{m-1}\lt y_m=d$$

The Riemann sum of a function f(x,y) over this partition of $$[a,b]\times [c,d]$$ is

$$\sum_{i=0}^{m}\sum_{j=0}^{n}f(u_i,v_j)\Delta x_i\Delta y_j$$

where $$(u_i,v_j)$$ is some point in the rectangle $$(x_{i-1},x_i)\times (y_{j-1},y_j) $$ and $$\Delta x_i=x_i-x_{i-1}, \Delta y_j=y_j-y_{j-1}$$

We then define the double integral of a function f(x,y) in the rectangular region $$[a,b]\times [c,d]$$ to be the limit of the Riemann sum as maximum values of $$\Delta x_i\ and \Delta y_j$$ and approach zero:

$$\iint_{[a,b]\times [c,d]}f(x,y)dA=\lim_{max\Delta x_i\to 0, max\Delta y_j\to 0}\sum_{i=0}^{m}\sum_{j=0}^{n}f(u_i,v_j)\Delta x_i\Delta y_j$$

To define the double integral over a bounded region R other than a rectangle, we choose a rectangle $$[a,b]\times [c,d]$$ that contains R and define the function g(x,y) so that

$$g(x,y)=f(x,y) if f(x,y)\in R$$
$$g(x,y)=0 if f(x,y)\notin R$$

Elementary General Region Of Integration In Double Integral
Elementary General Region Of Integration In Double Integral

Then the double integral of the function f(x,y) over a general region R is defined to be

$$\iint_{R}f(x,y)dA=\iint_{[a,b]\times [c,d]}g(x,y)dA$$

A Quick Refresh Of Area Under The Curve

From single variable calculus, we know that integrals let us compute the area under a curve. For example, the area under the graph of $$y=\frac{1}{4}x^2+1$$ between the values x = -3,  and x=3 is

$$\int_{-3}^{3}\bigl(\frac{1}{4}x^2+1\bigr)dx$$

A nice way to think about this is to imagine adding the areas of infinitely many, infinitely thin rectangles which sweep under the curve in the specified region:

You can think of the value of the function $$g(x) =\frac{1}{4}x^2+1$$ as being the height of each rectangle, dx, as being the infinitesimal width, and ∫ as being a pumped-up summing machine that’s able to handle the idea of infinitely many infinitely small things. Written more abstractly, this looks like

$$\int_{x_1}^{x_2}g(x)dx$$

Area Under A Volume

For our volume problem, we can do something similar. Our strategy will be to

  • Subdivide the volume into slices with two-dimensional areas
  • Compute the areas of these slices
  • Combine them all together to get the volume as a whole.

Think of two-dimensional slices of the volume under the graph of f(x, y). Specifically, take all the slices representing a constant value of y;

Consider just one of those slices, such as the one representing $$y=\frac{\pi}{2}$$ The area of that slice is given by the integral

$$\int_{0}^{2}f\bigl(x,\frac{\pi}{2}\bigr)dx =\int_{0}^{2}\bigl(x+sin\frac{\pi}{2}+1\bigr)dx$$

Written more abstractly, for a given value of y, the area of that slice is $$\int_{0}^{2}f(x,y)dx$$

Notice, this is an integral with respect to x, as indicated by the dx, so as far as the integral is concerned, the symbol “y” represents a constant.

When you perform this integral, it will be some expression of y.

Examples

$$Find\ the\ volume\ V\ under\ the\ plane\ z=8x+6y\ over\ the\ rectangle\ R=[0,1]\times[0,2]$$

We see that f(x,y)=8x+6y ≥ 0 for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2 , so:

$$V=\int_{0}^{2}\int_{0}^{1}(8x+6y)dx\ dy$$
$$V=\int_{0}^{2}\biggl([4x^2+6xy]_{x=0}^{x=1}\biggr) dy$$
$$V=\int_{0}^{2}\biggl(4+6y\biggr) dy$$
$$V=\biggl(4y+3y^2\biggr)_{y=0}^{y=2}$$
$$V=20$$

Suppose we had switched the order of integration. We can verify that we still get the same answer:

$$V=\int_{0}^{1}\int_{0}^{2}(8x+6y)dy\ dx$$
$$V=\int_{0}^{1}\biggl([8xy+3y^2]_{y=0}^{y=2}\biggr) dx$$
$$V=\int_{0}^{1}\biggl(16x+12\biggr) dx$$
$$V=\biggl(8x^2+12x\biggr)_{x=0}^{x=1}$$
$$V=20$$

Applications

The concept of double integrals arises in many fields of science and engineering, including the calculation of

  • A 2D region’s area
  • Defining the volume
  • Masses of 2D plates
  • Force on a two-dimensional plate
  • A function’s average
  • The center of mass and the moment of inertia
  • Area of the surface

FAQs

What is a double line integral?

A line integral is an integral where the function to be integrated is evaluated along a curve and a surface integral is a generalization of multiple integrals to integration over surfaces. It can be thought of as the double integral analog of the line integral.

What is the difference between double and triple integrals?

A double integral is used for integrating over a two-dimensional region, while a triple integral is used for integrating over a three-dimensional region.

What does the second integral tell us?

If a(t) is the acceleration at time t, then the first integral gives the velocity v(t), and the second integral gives the displacement s(t). That is the most important application, perhaps the only application, of integrating twice that you will meet this coming term.

Can double integrals be zero?

That double integral is telling you to sum up all the function values of x2−y2 over the unit circle. To get 0 here means that either the function does not exist in that region OR it’s perfectly symmetrical over it.

How do you do double integrals on a calculator?

The procedure to use the double integral calculator is as follows:

Step 1: Enter the function and the limits in the input field.
Step 2: Now click the button “Calculate” to get the value.
Step 3: Finally, the result of the double integral will be displayed in the new window.

What is the difference between line integral and double integral?

Double integrals, which are integrals over planar regions. Line or path integrals, which are integrals over curves.