- #1
VinnyCee
- 489
- 0
Here is the problem:
Find the average value of [tex]f\left(x, y\right) = x\;y[/tex] for the quarter circle [tex]x^2 + y^2 \le 1[/tex] in the first quadrant.
Here is what I have:
Average value equation is [tex]\frac{1}{Area\;of\;R} \iint_{R} f\left(x, y\right) dA[/tex]
[tex]f\left(x, y\right) = x\;y = \left(r\;\cos\theta\right)\left(r\;\sin\theta\right)[/tex]
The area of one quarter of a unti circle is [tex]\frac{\pi}{4}[/tex], right?
[tex]Average = \frac{4}{\pi}\;\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\;\left(r\;\cos\theta\right)\left(r\;\sin\theta\right)\;r\;dr\;d\theta = \frac{1}{2\pi}[/tex]
Is this correct?
Find the average value of [tex]f\left(x, y\right) = x\;y[/tex] for the quarter circle [tex]x^2 + y^2 \le 1[/tex] in the first quadrant.
Here is what I have:
Average value equation is [tex]\frac{1}{Area\;of\;R} \iint_{R} f\left(x, y\right) dA[/tex]
[tex]f\left(x, y\right) = x\;y = \left(r\;\cos\theta\right)\left(r\;\sin\theta\right)[/tex]
The area of one quarter of a unti circle is [tex]\frac{\pi}{4}[/tex], right?
[tex]Average = \frac{4}{\pi}\;\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\;\left(r\;\cos\theta\right)\left(r\;\sin\theta\right)\;r\;dr\;d\theta = \frac{1}{2\pi}[/tex]
Is this correct?