\(\iint \frac{x y}{\sqrt{1-y^{2}}} d x d y\) Over the positive quadrant of the circle x² + y² = 1 is __________.

Concept:

We are given the double integral over the positive quadrant of the circle \( x^2 + y^2 = 1 \):

\( \iint \frac{xy}{√{1 - y^2}} \, dx \, dy \)

Region of Integration:

The circle \( x^2 + y^2 = 1 \) is a unit circle. The positive quadrant implies:

• \( x \geq 0 \),
• \( y \geq 0 \),
• \( x^2 + y^2 \leq 1 \)

So, the limits become:

• y from 0 to 1
• for each y, x goes from 0 to √(1 - y²) 

Calculation:

\( \iint \frac{xy}{√{1 - y^2}} \, dx \, dy = \int_{0}^{1} \int_{0}^{√{1 - y^2}} \frac{xy}{√{1 - y^2}} \, dx \, dy \)

Take constants outside the inner integral:

\( = \int_{0}^{1} \frac{y}{√{1 - y^2}} \int_{0}^{√{1 - y^2}} x \, dx \, dy \)

Inner integral:

\( \int_{0}^{√{1 - y^2}} x \, dx = \frac{(1 - y^2)}{2} \)

So the expression becomes:

\( \int_{0}^{1} \frac{y}{√{1 - y^2}} \cdot \frac{1 - y^2}{2} \, dy = \frac{1}{2} \int_{0}^{1} y √{1 - y^2} \, dy \)

Use substitution: Let \( u = 1 - y^2 \Rightarrow du = -2y \, dy \Rightarrow y \, dy = -\frac{1}{2} du \)

Change limits: \( y = 0 \rightarrow u = 1 \), \( y = 1 \rightarrow u = 0 \)

So the integral becomes:

\( \frac{1}{2} \cdot \left( -\frac{1}{2} \int_{1}^{0} √{u} \, du \right) = \frac{1}{4} \int_{0}^{1} u^{1/2} \, du \)

Integrating:

\( = \frac{1}{4} \cdot \left[ \frac{u^{3/2}}{3/2} \right]_0^1 = \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{6} \)