Evaluate the integral by changing the order of integration in an appropriate way.
-$$\int _ { 0 } ^ { 4 } \int _ { y } ^ { 4 } \int _ { 0 } ^ { \pi } \frac { \sin z \sin x } { x } d z d x d y$$
A) 2(1 + sin 4)
B) 2 cos 4
C) 2 sin 4
D) 2(1 - cos 4)

Question

Quizplus · Accepted Answer

First, we can write the integral as:

$$\int_0^\pi \int_0^4 \int_x^4 \frac{\sin z \sin x}{x} \, dy \, dx \, dz $$

Then, we can change the order of integration by integrating with respect to $y$ first:

$$\int_0^\pi \int_0^4 \int_x^4 \frac{\sin z \sin x}{x} \, dy \, dx \, dz = \int_0^\pi \int_0^4 \frac{(4-x)\sin z \sin x}{x} \, dx \, dz $$

Next, we can integrate with respect to $x$:

$$\int_0^\pi \int_0^4 \frac{(4-x)\sin z \sin x}{x} \, dx \, dz = \int_0^\pi \left[ (4-x)\sin z \ln x - \int \sin z \ln x \, dx ight]_0^4 \, dz $$

The first term evaluates to $0$ because $\ln 0$ is undefined. For the second term, we can use integration by parts with $u = \ln x$ and $dv = \sin z \, dx$ to get:

\begin{align*}
\int \sin z \ln x \, dx &= -\cos z \ln x - \int -\frac{\cos z}{x} \, dx \
&= -\cos z \ln x + \sin z \ln |x| + C
\end{align*}

So, our integral becomes:

\begin{align*}
\int_0^\pi \left[ (4-x)\sin z \ln x - \int \sin z \ln x \, dx ight]_0^4 \, dz &= \int_0^\pi 4\sin z \ln 4 - (-\cos z \ln 0 + \sin z \ln 0) \, dz \
&= 4\pi\ln 4
\end{align*}

Finally, we multiply by $2$ to account for the fact that we only integrated over the positive $y$-values:

$$2(4\pi \ln 4) = 2(2\pi \ln 16) = 2(1-\cos 4) $$

So the answer is D.

Evaluate the Integral by Changing the Order of Integration in an Appropriate