244.T.15.5.11 Evaluate the triple integral

[karush]

$tiny{244.T.15.5.11}$
$\textsf{Evaluate the triple integral}\\$
\begin{align*}\displaystyle
I_{\tiny{11}}&=\int_{0}^{\pi/6}\int_{0}^{1}\int_{-2}^{3} y\sin{z}
\, d\textbf{x} \, d\textbf{y} \, d\textbf{z}\\
&=\int_{0}^{\pi/6}\int_{0}^{1}
\Biggr|xy\sin{z} \Biggr|_{-2}^{3}
\, d\textbf{y} \, d\textbf{z}\\
ans&=\color{red}{\frac{5(2-\sqrt{3})}{4}}
\end{align*}ok just want to see if these first steps are correct with xyz

 

[MarkFL]

I would begin by writing:

$$I=\int_{0}^{\pi/6}\sin(z)\int_{0}^{1}y\int_{-2}^{3}\,dx\,dy\,dz$$

As $$\int_{-2}^{3}\,dx=5$$ we then have:

$$I=5\int_{0}^{\pi/6}\sin(z)\int_{0}^{1}y\,dy\,dz$$

Now continue...:)

 

[karush]

So then continuing..

$$\begin{align*}\displaystyle
I_{11}&=\int_{0}^{\pi/6}\int_{0}^{1}\int_{-2}^{3} y\sin{z}
\, dx \, dy \, dz\\
&=5\int_{0}^{\pi/6}\sin(z)
\int_{0}^{1}y \,dy\,dz\\
&=5\int_{0}^{\pi/6}
\sin(z)\Biggr|\frac{y^2}{2}\Biggr|_{0}^{1} \,dz\\
&=\frac{5}{2}\int_{0}^{\pi/6}\sin(z)\,dz\\
&=\frac{5}{2}\Biggr|-\cos(z)\Biggr|_{z=0}^{z=\pi/6}
=\frac{5}{2}\Biggr[-\frac{\sqrt{3}}{2}-(-1)\Biggr]
=\color{red}{\frac{5(2-\sqrt{3})}{4}}
\end{align*}
$$

hopefully
any suggest?

 

[Greg]

Incorrect notation; otherwise correct. :)

 

[karush]

do you the vertical bars?

 

[Greg]

Ah, I see you fixed it!

Observe:

$$\left.-\frac{5}{2}\cos(z)\right|_{z=0}^{z=\pi/6}$$

Now you can use pure bars!

For something bigger:

$$\left.\left(\frac{x^2}{2}+3x+4\right)\right|^{4}_{x=0}$$