Evaluate the double integral $\iint _ { R } y ^ { 3 } \sin x \cos x d A$ , where $R = \left\{ ( x , y ) \mid 0 \leq x \leq \frac { \pi } { 2 } , 0 \leq y \leq 1 \right\}$ .
A) $\frac { 1 } { 4 }$

B) $\frac { 1 } { 2 }$
C) $\frac { \pi } { 4 }$
D) $\frac { 1 } { 8 }$
E) $\frac { 5 } { 4 }$
F) $\frac { 3 } { 2 }$

G) $\frac { \pi } { 2 }$

H)0

Question

Evaluate the double integral $\iint _ { R } y ^ { 3 } \sin x \cos x d A$ , where $R = \left\{ ( x , y ) \mid 0 \leq x \leq \frac { \pi } { 2 } , 0 \leq y \leq 1 \right\}$ .
A) $\frac { 1 } { 4 }$

B) $\frac { 1 } { 2 }$ 
C) $\frac { \pi } { 4 }$ 
D) $\frac { 1 } { 8 }$ 
E) $\frac { 5 } { 4 }$ 
F) $\frac { 3 } { 2 }$

G) $\frac { \pi } { 2 }$

H)0

Quizplus · Accepted Answer

The integral can be separated into two parts, integrating with respect to $y$ and $x$ separately due to the nature of the function. The integral with respect to $y$ yields $\frac{y^4}{4}$ evaluated from 0 to 1, which is $\frac{1}{4}$. The integral with respect to $x$ of $\sin x \cos x$ can be simplified using the double angle formula to $\frac{1}{2}\sin(2x)$, which evaluates to $\frac{1}{2}$ over the given interval. Multiplying these results gives $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.

Evaluate the Double Integral ∬Ry3sin⁡xcos⁡xdA\iint _ { R } y ^ { 3 } \sin x \cos x d A∬R​y3sinxcosxdA

Evaluate the Double Integral $\iint _ { R } y ^ { 3 } \sin x \cos x d A$