$\text { Evaluate the iterated integral } \int _ { 0 } ^ { 2 } \int _ { y ^ { 2 } } ^ { 4 } \sqrt { x } \sin ( 2 x ) d x d y \text { by switching the order of }$ integration. Round your answer to three decimal places.
A) $0.538$
B) $0.558$
C) $30.538$
D) $1.538$
E) $4.538$

Question

$\text { Evaluate the iterated integral } \int _ { 0 } ^ { 2 } \int _ { y ^ { 2 } } ^ { 4 } \sqrt { x } \sin ( 2 x ) d x d y \text { by switching the order of }$ integration. Round your answer to three decimal places. 
A) $0.538$
B) $0.558$
C) $30.538$
D) $1.538$
E) $4.538$

Quizplus · Accepted Answer

We have 
$$
\int _ { 0 } ^ { 2 } \int _ { y ^ { 2 } } ^ { 4 } \sqrt { x } \sin ( 2 x ) d x d y
$$
We can rewrite the region of integration as
$$
\begin{aligned} D & =\left\{ ( x , y ) : 0 \leq y \leq 2 , y ^ { 2 } \leq x \leq 4 ight\} \ & =\left\{ ( x , y ) : x \in \left[ y ^ { 2 } , 4 ight] , y \in \left[ 0,2 ight] ight\} \end{aligned}
$$
So we can change the order of integration and write 
\begin{align}
& \int _ { 0 } ^ { 2 } \int _ { y ^ { 2 } } ^ { 4 } \sqrt { x } \sin ( 2 x ) d x d y \
& =\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { \sqrt { x } } \sqrt { x } \sin ( 2 x ) d y d x \
& =\int _ { 0 } ^ { 4 } \sqrt { x } \sin ( 2 x ) \int _ { 0 } ^ { \sqrt { x } } d y d x \
& =\int _ { 0 } ^ { 4 } \sqrt { x } \sin ( 2 x ) \sqrt { x } d x \
& =\int _ { 0 } ^ { 4 } x \sin ( 2 x ) d x \
& =\frac { 1 } { 2 } \int _ { 0 } ^ { 8 } \frac { 1 } { 4 } \sin u \, du & (u=2x)\
& =\frac { 1 } { 8 } \left[ - \cos 2 x ight] _ { 0 } ^ { 8 / 2 } \
& =\frac { 1 } { 8 } ( 1 - \cos 8 ) \
& \approx 0.538
\end{align}

Integration 0.5380.5380.538 B) 0.5580.5580.558 C) 30.53830.53830.538 D) 1.5381.5381.538 E) 4.5384.5384.538

Integration $0.538$ B) $0.558$ C) $30.538$ D) $1.538$ E) $4.538$