15.4.20 volumn via triple integrals

In summary, the region in the first octant bounded by the coordinate planes and the surface $z=4-x^2-y$ can be represented by the triple integral $\iiint\limits_{E} \, dzdydx$ with limits $0 \le z \le 4-x^2-y$, $0 \le y \le 4-x^2$, and $0 \le x \le z$. It can also be represented by the triple integral $\iiint\limits_{E} \, dydzdx$ with limits $0 \le y \le 4-x^2$, $0 \le z \le 4-x^2-y$, and $0 \le x \le z$.
  • #1
karush
Gold Member
MHB
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$\textsf{The region in the first octant bounded by the coordinate planes and the surface }$
View attachment 8112
$$z=4-x^2-y$$
$\textit{From the given equation we get}$
\begin{align*}\displaystyle
&0 \le z \le 4-x^2-y\\
&0 \le y \le 4-x^2\\
&0 \le x \le z
\end{align*}
$\textit{Thus the triple Integral results:}$
\begin{align*}\displaystyle
V&=\iiint\limits_{E} \, dzdyd
\quad =\int_{0}^{2}
\int_{0}^{4-x^2}
\int_{0}^{4-x^2-y}
\, dzdydx
\end{align*}
So just seeing if this is going the right direction
$\tiny{15.4.20}$
 
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  • #2
karush said:
$\textsf{The region in the first octant bounded by the coordinate planes and the surface }$

$$z=4-x^2-y$$
$\textit{From the given equation we get}$
\begin{align*}\displaystyle
&0 \le z \le 4-x^2-y\\
&0 \le y \le 4-x^2\\
&0 \le x \le z
\end{align*}
$\textit{Thus the triple Integral results:}$
\begin{align*}\displaystyle
V&=\iiint\limits_{E} \, dzdyd
\quad =\int_{0}^{2}
\int_{0}^{4-x^2}
\int_{0}^{4-x^2-y}
\, dzdydx
\end{align*}
So just seeing if this is going the right direction
$\tiny{15.4.20}$

It's a good thing you didn't need that equation on the upper left.

Okay, now do it all over dydzdx and dydxdz.
 

FAQ: 15.4.20 volumn via triple integrals

1. What is the concept of "15.4.20 volume via triple integrals"?

"15.4.20 volume via triple integrals" refers to a mathematical concept used in multivariable calculus to calculate the volume of a three-dimensional shape. It involves breaking down a shape into infinitesimally small pieces and integrating their volumes to find the total volume of the shape.

2. How is the volume of a three-dimensional shape calculated using triple integrals?

The volume of a three-dimensional shape is calculated by integrating the function representing the shape over three variables - x, y, and z. This is known as a triple integral and the resulting value is the volume of the shape.

3. What are the steps involved in using triple integrals to calculate volume?

The steps involved in using triple integrals to calculate volume include identifying the limits of integration for each variable, setting up the triple integral using the function representing the shape, solving the integral using standard integration techniques, and finally evaluating the integral to find the volume of the shape.

4. Can triple integrals be used to find the volume of any three-dimensional shape?

Yes, triple integrals can be used to find the volume of any three-dimensional shape as long as the shape can be represented by a function. This method is particularly useful for finding the volumes of irregular shapes or shapes with varying densities.

5. Are there any real-world applications of triple integrals in calculating volume?

Yes, triple integrals have many real-world applications in fields such as physics, engineering, and computer graphics. For example, they can be used to calculate the volume of an irregularly shaped object in physics experiments or to determine the amount of material needed to construct a complex structure in engineering.

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