*15.4.17 Volume between a cone and a sphere

[karush]

$\tiny{15.4.17}$
Find the volume of the given solid region
bounded below by the cone $z=\sqrt{x^2+y^2}$
and bounded above by the sphere $x^2+y^2+z^2=128$
using triple integrals

$\displaystyle\int_{a}^{b}\int_{c}^{d} \int_{e}^{f} \,dx\,dy \,dz$

not real sure where to to start with this?

 

[Prove It]

Re: 15.4.17 volume between a cone and a sphere

$\tiny{15.4.17}$
Find the volume of the given solid region
bounded below by the cone $z=\sqrt{x^2+y^2}$
and bounded above by the sphere $x^2+y^2+z^2=128$
using triple integrals

$\displaystyle\int_{a}^{b}\int_{c}^{d} \int_{e}^{f} \,dx\,dy \,dz$

not real sure where to to start with this?

Bounded ABOVE by $\displaystyle \begin{align*} x^2 + y^2 + z^2 = 128 \implies z = \sqrt{128 - \left( x^2 + y^2 \right) } \end{align*}$ and bounded BELOW by $\displaystyle \begin{align*} z = \sqrt{x^2 + y^2} \end{align*}$ means $\displaystyle \begin{align*} \sqrt{ x^2 + y^2 } \leq z \leq \sqrt{ 128 - \left( x^2 + y^2 \right) } \end{align*}$.

Now we need to determine how far out we go in each direction in the x-y plane. Notice that each cross-section parallel to the x-y plane is a full circle, and the maximum circle will be where the cone and sphere intersect, so where:

$\displaystyle \begin{align*} 128 - \left( x^2 + y^2 \right) &= x^2 + y^2 \\ 128 &= 2 \left( x^2 + y^2 \right) \\ 64 &= x^2 + y^2 \end{align*}$

So the maximum circle is of radius 8 units centred at the origin.

It would be easiest to do this triple integral in cylindrical polar co-ordinates, so with $\displaystyle \begin{align*} 0 \leq r \leq 8 \end{align*}$ and $\displaystyle \begin{align*} 0 \leq \theta \leq 2\,\pi \end{align*}$, so your z boundaries are $\displaystyle \begin{align*} r \leq z \leq \sqrt{128 - r^2} \end{align*}$ and your triple integral is

$\displaystyle \begin{align*} V &= \int_0^{2\,\pi}{\int_0^8{\int_r^{\sqrt{128 - r^2}} r\,\mathrm{d}z \,\mathrm{d}r }\,\mathrm{d}\theta} \end{align*}$

 

[karush]

Re: 15.4.17 volume between a cone and a sphere

thanks that was really a great help

much mahalo...