What is the value of the triple integral for the given limits and function?

[karush]

\begin{align}\displaystyle
v_{\tiny{s6.15.6.3}}&=\displaystyle
\int_{0}^{1}\int_{0}^{z}\int_{0}^{x+z}
6xz \quad
\, dy \, dx\, dz
\end{align}

$\text{ok i kinda got ? with $x+z$ to do the first step?}\\$
$\text{didn't see an example}$

 

[MarkFL]

I would write:

$$I=6\int_0^1 z\int_0^z x\left(\int_0^{x+z}\,dy\right)\,dx\,dz$$

What would your next step be?

 

[karush]

actually i was going to that,,,
but to scared
ill get to this in the morn.
to hard with just a small cell phone

 

[karush]

I would write:

$$I=6\int_0^1 z\int_0^z x\left(\int_0^{x+z}\,dy\right)\,dx\,dz$$

What would your next step be?

\begin{align}\displaystyle
I&=6\int_0^1 z\int_0^z x\left[\int_0^{x+z}\,dy\right]\,dx\,dz\\
&=6\int_0^1 z\int_0^z x\Biggr|y\Biggr|_0^{x+z}\,dy \, dx \, dz\\
&=6\int_0^1 z\int_0^z x(x+z) \,dx\,dz\\
&=6\int_0^1 z\int_0^z (x^2+xz) \,dx\,dz\\
\end{align}
hopefully

 

[topsquark]

\begin{align}\displaystyle
I&=6\int_0^1 z\int_0^z x\left[\int_0^{x+z}\,dy\right]\,dx\,dz\\
&=6\int_0^1 z\int_0^z x\Biggr|y\Biggr|_0^{x+z}\,dy \, dx \, dz\\
&=6\int_0^1 z\int_0^z x(x+z) \,dx\,dz\\
&=6\int_0^1 z\int_0^z (x^2+xz) \,dx\,dz\\
\end{align}
hopefully

So far so good. What's the next step? (Remember that z is treated as a constant when you do the x integration.)

-Dan

 

[karush]

So far so good. What's the next step? (Remember that z is treated as a constant when you do the x integration.)

-Dan

\begin{align*}\displaystyle
&=6\int_0^1 z\int_0^z (x^2+xz) \,dx\,dz\\
&=6z\int_0^1 \biggr|\frac{x^3}{3}+\frac{x^2}{2}z\biggr|_0^z \,dz\\
\end{align*}

no sure about $z$ ?

 

[topsquark]

\begin{align*}\displaystyle
&=6\int_0^1 z\int_0^z (x^2+xz) \,dx\,dz\\
&=6z\int_0^1 \biggr|\frac{x^3}{3}+\frac{x^2}{2}z\biggr|_0^z \,dz\\
\end{align*}

no sure about $z$ ?

Again, so far so good. Now plug in the limits for x. You will be substituting x = z and x = 0 into your limit.

[math]\left ( \frac{z^3}{3}+\frac{z^2}{2}z \right ) - \left ( \frac{0^3}{3}+\frac{0^2}{2}z \right )[/math]

-Dan

Edit: Whoops! There is an error in your work but it doesn't involve the x integration you just did. That z on the left side of the integral needs to be back under the integration symbol. z is not considered to be a constant when you are integrating over z. After you integrate over x you should have
[math]6 \int_0^1 z \cdot \left [ \left ( \frac{z^3}{3}+\frac{z^2}{2}z \right ) - \left ( \frac{0^3}{3}+\frac{0^2}{2}z \right ) \right ]
~ dz[/math]

 

[karush]

Again, so far so good. Now plug in the limits for x. You will be substituting x = z and x = 0 into your limit.

[math]\left ( \frac{z^3}{3}+\frac{z^2}{2}z \right ) - \left ( \frac{0^3}{3}+\frac{0^2}{2}z \right )[/math]
[math]6 \int_0^1 z \cdot \left [ \left ( \frac{z^3}{3}+\frac{z^2}{2}z \right ) - \left ( \frac{0^3}{3}+\frac{0^2}{2}z \right ) \right ]
~ dz[/math]

\begin{align*}\displaystyle
&I=6 \int_0^1 z \cdot \left [ \left ( \frac{z^3}{3}+\frac{z^2}{2}z \right )
- \left ( \frac{0^3}{3}+\frac{0^2}{2}z \right ) \right ] ~ dz\\
&=6\int_{0}^{1}\frac{z^4}{3}+\frac{z^4}{2} \,dz\\
&=6\int_{0}^{1}\frac{5z^4}{6} \,dz\\
&=5\left[\frac{z^5}{5}\right]_0^1 \\
&=\color{red}{1}
\end{align*}View attachment 7869