Average Value Limits

[M98Ranger]

I am having difficulty figuring out the limits of integration for this double integral problem. The problem is worded like this;

Find the average value of the function y=x*y over the quarter circle \(\displaystyle \
x^2 + y^2 \le 1
\\)
in the first quadrant.

Would it be \(\displaystyle \
0 \le x \le 1\,and\,0 \le y \le 1
\\)? Would the integral be set up like this? \(\displaystyle \
\int_0^1 {\int_0^1 {xydxdy} }
\\) *\(\displaystyle \
\frac{1}{{\int_0^1 {\int_0^1 {dxdy} } }}
\\)? Any help would be appreciated very much. For those that have forgotten the average value of a function f over a Region is \(\displaystyle \
\frac{1}{{area\;of\;R}}\; \times \;\int\limits_R {\int {fdA} }
\\)

 

[tkhunny]

No, that's a square.

Try one of them as you have it and the other as a function of the first. Since we are in the first quadrant, \(\displaystyle y = \sqrt{1-x^{2}}\)

 

[M98Ranger]

Well here is what I got, however it doesn't feel right because it is symbolic and I thought that I should come out with a numerical result. Here is what I have.

\(\displaystyle \
\begin{array}{l}
Area = \int_0^1 {\int_0^{\sqrt {1 - x^2 } } {1dxdy} } \to \frac{1}{{Area}} = \frac{1}{{\sqrt {1 - x^2 } }} \to \int_0^1 {\int_0^{\sqrt {1 - x^2 } } {xy\,dxdy} } = \frac{{1 - x^2 }}{4} \\
\left( {\frac{{1 - x^2 }}{4}} \right)\left( {\frac{1}{{\sqrt {1 - x^2 } }}} \right) = \frac{{\sqrt {1 - x^2 } }}{4} \\
\end{array}
\\) \(\displaystyle \
\frac{{\sqrt {1 - x^2 } }}{4} = Average\;Value
\\) ?? Is this correct then?
Did I follow what you were saying? I kind of understand average value, but with this problem I am not able to visualize what exactly I am supposed to be doing.

 

[daon]

Well here is what I got, however it doesn't feel right because it is symbolic and I thought that I should come out with a numerical result. Here is what I have.

\(\displaystyle \
\begin{array}{l}
Area = \int_0^1 {\int_0^{\sqrt {1 - x^2 } } {1dxdy} } \to \frac{1}{{Area}} = \frac{1}{{\sqrt {1 - x^2 } }} \to \int_0^1 {\int_0^{\sqrt {1 - x^2 } } {xy\,dxdy} } = \frac{{1 - x^2 }}{4} \\
\left( {\frac{{1 - x^2 }}{4}} \right)\left( {\frac{1}{{\sqrt {1 - x^2 } }}} \right) = \frac{{\sqrt {1 - x^2 } }}{4} \\
\end{array}
\\) \(\displaystyle \
\frac{{\sqrt {1 - x^2 } }}{4} = Average\;Value
\\) ?? Is this correct then?
Did I follow what you were saying? I kind of understand average value, but with this problem I am not able to visualize what exactly I am supposed to be doing.

Notice you are plugging in a function of x as a limit for the integration WRT x. You should have NO x's left after your first integral.

Either: 1) use dA = dydx, or 2) change your integration limits

Also: The area of your R is \(\displaystyle \frac{\pi}{4}\)

 

[M98Ranger]

Thankyou Very Much. I just realized that I gotmy order of integration mixed up when I woke up this morning.