Steps for Setting Up Triple Integrals

[harpazo]

I am really struggling setting up triple integrals. I need steps, simple steps normally applied when setting up integrals given a specific region.

 

[Prove It]

I am really struggling setting up triple integrals. I need steps, simple steps normally applied when setting up integrals given a specific region.

Well the first step is always drawing a sketch of your region...

 

[harpazo]

Well the first step is always drawing a sketch of your region...

Ok but can you provide steps? Find the bounds has been a big challenge for me. I can easily integrate when the set up is given.

 

[MarkFL]

Let's look at a very simple example. Consider the region bounded by:

\(\displaystyle z=0\)

\(\displaystyle z=h\)

\(\displaystyle x^2+y^2=r^2\)

When we make a sketch of the region, we see is is a right circular cylinder of radius $r$ and height $h$, so we know its volume is $V=\pi r^2h$. To set up a triple integral to compute the volume, we see that:

\(\displaystyle 0\le z\le h\)

\(\displaystyle -r\le y\le r\)

\(\displaystyle -\sqrt{r^2-y^2}\le x\le\sqrt{r^2-y^2}\)

And so we may express the volume as:

\(\displaystyle V=\int_0^h\int_{-r}^{r}\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\,dx\,dy\,dz\)

Using symmetry, we may state:

\(\displaystyle V=4\int_0^h\int_{0}^{r}\int_{0}^{\sqrt{r^2-y^2}}\,dx\,dy\,dz\)

Evaluating, we have:

\(\displaystyle V=4\int_0^h\int_{0}^{r}\sqrt{r^2-y^2}\,dy\,dz=\pi r^2\int_0^h\,dz=\pi r^2h\quad\checkmark\)

Likewise, a sphere of radius $r$ can be set up as follows:

\(\displaystyle V=8\int_0^r\int_0^{\sqrt{r^2-z^2}}\int_0^{\sqrt{r^2-y^2}}\,dx\,dy\,dz\)

Draw a sketch of a sphere centered at the origin of radius $r$ and see if you can confirm these limits. :D

 

[harpazo]

Let's look at a very simple example. Consider the region bounded by:

\(\displaystyle z=0\)

\(\displaystyle z=h\)

\(\displaystyle x^2+y^2=r^2\)

When we make a sketch of the region, we see is is a right circular cylinder of radius $r$ and height $h$, so we know its volume is $V=\pi r^2h$. To set up a triple integral to compute the volume, we see that:

\(\displaystyle 0\le z\le h\)

\(\displaystyle -r\le y\le r\)

\(\displaystyle -\sqrt{r^2-y^2}\le x\le\sqrt{r^2-y^2}\)

And so we may express the volume as:

\(\displaystyle V=\int_0^h\int_{-r}^{r}\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\,dx\,dy\,dz\)

Using symmetry, we may state:

\(\displaystyle V=4\int_0^h\int_{0}^{r}\int_{0}^{\sqrt{r^2-y^2}}\,dx\,dy\,dz\)

Evaluating, we have:

\(\displaystyle V=4\int_0^h\int_{0}^{r}\sqrt{r^2-y^2}\,dy\,dz=\pi r^2\int_0^h\,dz=\pi r^2h\quad\checkmark\)

Likewise, a sphere of radius $r$ can be set up as follows:

\(\displaystyle V=8\int_0^r\int_0^{\sqrt{r^2-z^2}}\int_0^{\sqrt{r^2-y^2}}\,dx\,dy\,dz\)

Draw a sketch of a sphere centered at the origin of radius $r$ and see if you can confirm these limits. :D

An excellent explanation. Thanks.