MHB 244.T.15.5.11 Evaluate the triple integral

karush
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$tiny{244.T.15.5.11}$
$\textsf{Evaluate the triple integral}\\$
\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
 
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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...:)
 
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?
 
Last edited:
Incorrect notation; otherwise correct. :)
 
do you the vertical bars?
 
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:

$$\left.\left(\frac{x^2}{2}+3x+4\right)\right|^{4}_{x=0}$$
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

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