# 232.5a Evaluate the double integral

• MHB
• karush
In summary, the double integral is evaluated by first separating the x and y variables and then evaluating the single integral for both x = 2 and x = 1. This results in the integral being simplified to $\int\limits_{-1}^{1}2y\sqrt{4+y^2}\;dy$ and $\int\limits_{-1}^{1}y\sqrt{1+y^2}\;dy$.
karush
Gold Member
MHB
$\tiny{232.5a}\\ \textsf{Evaluate the double integral}$
\begin{align*}\displaystyle
I_a&=\iint\limits_{R} xy\sqrt{x^2+y^2} \, dA \\
R&=[0,2]\times[-1,1]
\end{align*}
Ok, just want to see if I made the first step correct.
this looks like simply a rectangle so x and y are basically interchangeable
\begin{align*}
&=\int_{-1}^{1} \int_{0}^{2} xy\sqrt{x^2+y^2} \,dx \,dy
\end{align*}
however the next step looks kinda daunting

How about the substitution $x=y\tan\theta$?

greg1313 said:
How about the substitution $x=y\tan\theta$?

are you suggesting that

\begin{align*}
&=\int_{-1}^{1} \int_{a}^{b} y^2\tan\theta \, \sqrt{y^2\tan^2\theta+y^2} \,d\theta \,dy
\end{align*}

not sure what a and b would be on the limits

karush said:
$\tiny{232.5a}\\ \textsf{Evaluate the double integral}$
\begin{align*}\displaystyle
I_a&=\iint\limits_{R} xy\sqrt{x^2+y^2} \, dA \\
R&=[0,2]\times[-1,1]
\end{align*}
Ok, just want to see if I made the first step correct.
this looks like simply a rectangle so x and y are basically interchangeable
\begin{align*}
&=\int_{-1}^{1} \int_{0}^{2} xy\sqrt{x^2+y^2} \,dx \,dy
\end{align*}
however the next step looks kinda daunting

Ummm... No. Do an experiment for me. Take x = 2 and evaluate the single integral. Then, back up and take x = 1 and evaluate the single integral. See if you notice something. This is an eyeball problem.

Definition: "Eyeball Problem" - Just look at it and call out the answer before your neighbor beats you to it.

that is giving me a headache

I presume you mean separate x and y

karush said:
that is giving me a headache

I presume you mean separate x and y

No, I mean just pick x = 2 and then ignore that x exists and ignore the inner integral.

$\int\limits_{-1}^{1}2y\sqrt{4+y^2}\;dy$

What do you get?

Now, x = 1.

$\int\limits_{-1}^{1}y\sqrt{1+y^2}\;dy$

What do you get?

Last edited:

## 1. What is a double integral?

A double integral is a type of mathematical operation that involves calculating the area under a surface in two-dimensional space. It is used in calculus and is an extension of the concept of a single integral.

## 2. How is a double integral evaluated?

A double integral is evaluated by dividing the area into small rectangles and calculating the sum of the areas of these rectangles. This sum can be expressed as a definite integral, which can then be solved using integration techniques.

## 3. What is the difference between a double integral and a single integral?

The main difference between a double integral and a single integral is that a single integral calculates the area under a one-dimensional curve, while a double integral calculates the area under a two-dimensional surface.

## 4. What is the purpose of evaluating a double integral?

Evaluating a double integral is useful in solving various problems in physics, engineering, and other fields that involve calculating the volume, mass, or other properties of a three-dimensional object or surface.

## 5. Can a double integral be evaluated using software?

Yes, there are many software programs, such as Mathematica, Matlab, and Wolfram Alpha, that can evaluate double integrals numerically or symbolically. However, it is important to understand the concepts and techniques behind double integration before relying on software for solutions.

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