If R is the region bounded by $x = \sqrt { 1 - y ^ { 2 } } , y = x$ and the positive x-axis, then $\iint$_{R $x d A$} in polar form is A) $\int _ { 0 } ^ { \frac { \pi } { 4 } } \left[ \int _ { 0 } ^ { 1 } r \cos \theta d r \right] d \theta$ B) $\int _ { 0 } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 1 } r \cos \theta d r \right] d \theta$ C) $\int _ { 0 } ^ { \frac { \pi } { 4 } } \left[ \int _ { 0 } ^ { 1 } r ^ { 2 } \cos \theta d r \right] d \theta$ D) $\int _ { 0 } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 1 } r ^ { 2 } \cos \theta d r \right] d \theta$ E) $\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 1 } r ^ { 2 } \cos \theta d r \right] d \theta$

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The region R is bounded by the curve $x = \sqrt{1 - y^2}$, the line $y = x$, and the positive x-axis. In polar coordinates, $x = r\cos\theta$ and $y = r\sin\theta$. The curve $x = \sqrt{1 - y^2}$ represents a circle of radius 1 centered at the origin, and the line $y = x$ corresponds to $\theta = \frac{\pi}{4}$. The integral of $x dA$ in polar coordinates involves $r^2\cos\theta dr d\theta$, and the limits for $\theta$ are from 0 to $\frac{\pi}{4}$ to cover the region R, making the correct choice $\int _ { 0 } ^ { \frac { \pi } { 4 } } \left[ \int _ { 0 } ^ { 1 } r ^ { 2 } \cos \theta d r \right] d \theta$.

If R Is the Region Bounded By x=1−y2,y=xx = \sqrt { 1 - y ^ { 2 } } , y = xx=1−y2​,y=x

If R Is the Region Bounded By $x = \sqrt { 1 - y ^ { 2 } } , y = x$