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