# 213.15.4.17 triple integral of bounded by cone and sphere

• MHB
• karush
In summary, the volume of the given solid region bounded by the cone is $128$ and bounded above by the sphere.
karush
Gold Member
MHB
$\textsf{Find the volume of the given solid region bounded by the cone}$
$$\displaystyle z=\sqrt{x^2+y^2}$$
$\textsf{and bounded above by the sphere}$
$$\displaystyle x^2+y^2+z^2=128$$
$\textsf{ using triple integrals}$

\begin{align*}\displaystyle
V&=\iiint\limits_{R}p(x,y,z) \, dV
\end{align*}

ok I don't think I can get the Integral set up the way the equations are written
I assume we break it up into $x=, y=$ and $z=$

\begin{align*}\displaystyle
z_{cone}&=\sqrt{x^2+y^2}\\
z^2&=x^2+y^2\\
x^2&=z^2-y^2\\
\therefore x_{cone}&=\sqrt{z^2-y^2}\\
y^2&=z^2-x^2\\
\therefore y_{cone}&=\sqrt{z^2-x^2}
\end{align*}.

Last edited:
Because of the symmetry here, I think I would be inclined to use cylindrical coordinates.

In cylindrical coordinates, $$\displaystyle x= r cos(\theta)$$, $$\displaystyle y= r sin(\theta)$$, and $$\displaystyle z= z$$. The cone becomes z= r and the sphere becomes $$\displaystyle r^2+ z^2= 128$$. The upper portion of the sphere, above the cone, is given by $$\displaystyle z= \sqrt{128- r^2}$$. The cone and sphere meet where $$\displaystyle r^2+ r^2= 2r^2= 128$$ so $$\displaystyle r= 8$$. To cover the region between the cone and the sphere, $$\displaystyle \theta$$ must go from 0 to $$\displaystyle 2\pi$$. r must go from 0 to 8, and, for each r, z goes from r, at the cone, to $$\displaystyle \sqrt{128- r^2}$$, at the sphere. The "differential of volume" in cylindrical coordinates is $$\displaystyle r drd\theta dz$$ so the integral you want is

$$\displaystyle \int_0^{2\pi}\int_0^8 \int_r^{\sqrt{128- r^2}} r dz drd\theta$$.

\begin{align*}\displaystyle
V&=\int_0^{2\pi}\int_0^8 \int_r^{\sqrt{128- r^2}} r dz dr d\theta\\
&=\int_0^{2\pi}\int_0^8\biggr[r^2\biggr]_r^{\sqrt{128- r^2}} dz \, d\theta\\
\end{align*}uhh the $dr$ ?

karush said:
\begin{align*}\displaystyle
V&=\int_0^{2\pi}\int_0^8 \int_r^{\sqrt{128- r^2}} r dz dr d\theta\\
&=\int_0^{2\pi}\int_0^8\biggr[r^2\biggr]_r^{\sqrt{128- r^2}} dz \, d\theta\\
\end{align*}uhh the $dr$ ?
uhh, what did you do with it? The first (inner) integral is with respect to z. When you have done that integral, it is the "dz" that disappears, not "dr".
$$\displaystyle \int_r^{\sqrt{128- r^2}} r dz= \left[rz\right]_r^{\sqrt{128- r^2}}=r\sqrt{128- r^2}- r^2$$

So the next integral is $$\displaystyle \int_0^8 r\sqrt{128- r^2}- r^2 dr$$.

Of course, $$\displaystyle \int_0^{2\pi} d\theta= 2\pi$$ and just multiplies the other integral.

(And, by the way, the integral of "$$\displaystyle rdr$$" is NOT "$$\displaystyle r^2$$".)

## 1. What is the purpose of calculating a triple integral of a bounded region?

The purpose of calculating a triple integral of a bounded region is to determine the volume of the region. This can be useful in various applications, such as calculating the mass of an object with varying density or finding the center of mass of a three-dimensional object.

## 2. What does "bounded by cone and sphere" mean in the context of a triple integral?

When a region is bounded by a cone and sphere, it means that the boundaries of the region are defined by these two three-dimensional shapes. In other words, the region is enclosed by a cone and a sphere, creating a specific shape that can be represented mathematically.

## 3. How is the triple integral of a bounded region calculated?

The triple integral of a bounded region is calculated by breaking the region into infinitesimally small parts and summing up the volume of each part. This is represented mathematically as ∭f(x,y,z) dV, where f(x,y,z) represents the function that determines the volume of each part and dV represents the infinitesimal volume element.

## 4. What is the significance of the numbers "213.15.4.17" in the context of a triple integral?

The numbers "213.15.4.17" do not have any significance in the context of a triple integral. They are most likely used as a placeholder or example IP address in the question.

## 5. Are there any practical applications of calculating a triple integral of a bounded region?

Yes, there are many practical applications of calculating a triple integral of a bounded region, such as in physics, engineering, and computer graphics. For example, it can be used to calculate the volume of a three-dimensional object, the work done by a force acting on a three-dimensional object, or the amount of fluid flowing through a pipe with varying cross-sectional area.

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