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

In summary, the equation $I=6\int_0^1 z\int_0^z x\left(\int_0^{x+z}\,dy\right)\,dx\,dz$ can be solved for z when x is substituted into the limits.
  • #1
karush
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\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?}\\$
$\text{didn't see an example}$
 
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  • #2
I would write:

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

What would your next step be?
 
  • #3
actually i was going to that,,,
but to scared
ill get to this in the morn.
to hard with just a small cell phone
 
  • #4
MarkFL said:
I would write:

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

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
 
  • #5
karush said:
\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
 
  • #6
topsquark said:
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 $z$ ?
 
  • #7
karush said:
\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.

\(\displaystyle \left ( \frac{z^3}{3}+\frac{z^2}{2}z \right ) - \left ( \frac{0^3}{3}+\frac{0^2}{2}z \right )\)

-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
\(\displaystyle 6 \int_0^1 z \cdot \left [ \left ( \frac{z^3}{3}+\frac{z^2}{2}z \right ) - \left ( \frac{0^3}{3}+\frac{0^2}{2}z \right ) \right ]
~ dz\)
 
  • #8
topsquark said:
Again, so far so good. Now plug in the limits for x. You will be substituting x = z and x = 0 into your limit.

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

\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
 

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

What is a triple integral?

A triple integral is an extension of the concept of a single or double integral in calculus. It involves integrating a function of three variables over a three-dimensional region and is used to calculate the volume of a solid or the mass of a three-dimensional object.

How do you set up a triple integral?

To set up a triple integral, you need to determine the limits of integration for each variable. This is done by defining the boundaries of the three-dimensional region in terms of the variables, and then integrating the function over those boundaries. The order of integration (which variable to integrate first) is also important and can be determined by considering the shape of the region and the function being integrated.

What is the difference between a triple integral and a double integral?

The main difference between a triple integral and a double integral is the number of variables involved. A double integral involves integrating a function over a two-dimensional region, while a triple integral involves integrating a function over a three-dimensional region. In practical terms, this means that a triple integral is used to calculate the volume or mass of a three-dimensional object, while a double integral is used for two-dimensional objects.

What are some applications of triple integrals?

Triple integrals have many applications in physics, engineering, and other fields. They can be used to calculate the mass or volume of a three-dimensional object, as well as to determine the average value of a function over a three-dimensional region. They are also used in the calculation of moments of inertia, which is important in mechanics and engineering.

How do you solve a triple integral?

Solving a triple integral involves setting up the integral, determining the limits of integration, and then evaluating the integral using techniques such as substitution or integration by parts. In some cases, the integral may need to be approximated using numerical methods. It is also important to carefully consider the order of integration to ensure an accurate solution. Practice and familiarity with mathematical concepts and techniques are key to successfully solving triple integrals.

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