Evaluate the double integral $\iint _ { R } x d A$ , where $R = \left\{ ( x , y ) \mid y - 1 \leq x \leq \sqrt { 1 - y ^ { 2 } } , 0 \leq y \leq 1 \right\}$ .
A) $\frac { 1 } { 6 }$

B) $\frac { 1 } { 4 }$
C) $\frac { 1 } { 3 }$
D) $\frac { 1 } { 2 }$
E) $\frac { 2 } { 3 }$
F) $\frac { 3 } { 4 }$

G) $\frac { 5 } { 6 }$
H)1

Question

Evaluate the double integral $\iint _ { R } x d A$ , where $R = \left\{ ( x , y ) \mid y - 1 \leq x \leq \sqrt { 1 - y ^ { 2 } } , 0 \leq y \leq 1 \right\}$ .
A) $\frac { 1 } { 6 }$

B) $\frac { 1 } { 4 }$ 
C) $\frac { 1 } { 3 }$ 
D) $\frac { 1 } { 2 }$ 
E) $\frac { 2 } { 3 }$ 
F) $\frac { 3 } { 4 }$

G) $\frac { 5 } { 6 }$ 
H)1

Quizplus · Accepted Answer

To evaluate the given double integral, we first need to understand the limits of integration from the given region $R$. The region is bounded by $y - 1 \leq x$, $x \leq \sqrt{1 - y^2}$, and $0 \leq y \leq 1$. The integral becomes:$$\int_{0}^{1} \int_{y-1}^{\sqrt{1-y^2}} x \, dx \, dy$$First, integrate with respect to $x$:$$\int_{0}^{1} \left[ \frac{1}{2}x^2 \right]_{y-1}^{\sqrt{1-y^2}} dy = \int_{0}^{1} \left( \frac{1}{2}(\sqrt{1-y^2})^2 - \frac{1}{2}(y-1)^2 \right) dy$$$$= \int_{0}^{1} \left( \frac{1}{2}(1-y^2) - \frac{1}{2}(y^2 - 2y + 1) \right) dy$$$$= \int_{0}^{1} \left( \frac{1}{2} - y + \frac{1}{2}y^2 \right) dy$$$$= \left[ \frac{1}{2}y - \frac{1}{2}y^2 + \frac{1}{6}y^3 \right]_{0}^{1}$$$$= \frac{1}{2} - \frac{1}{2} + \frac{1}{6} = \frac{1}{6}$$Therefore, the correct answer is $\frac{1}{6}$, which corresponds to choice A.

Evaluate the Double Integral ∬RxdA\iint _ { R } x d A∬R​xdA

Evaluate the Double Integral $\iint _ { R } x d A$