Set up triple integral

[Daniel_Feldman]

I need to set up a triple integral to find the volume of the region in octant 1 bounded by x=0, y=0, below by z=2, and above by z=4-x^2-y^2.


I figured my limits for x would be 0 to 2, for y 0 to 2, and for z 2 to4-x^2-y^2, but that is really messy and complicated. Should I use spherical coords, and if so, what would be the best way to do that?

 

[Daniel_Feldman]

Maybe z can be rewritten as 4-(x^2+y^2)=4-r^2m and then I can apply cylindricals...but I'm still not sure.

 

[soroban]

Hello, Daniel!

I need to set up a triple integral to find the volume of the region in octant 1
bounded by: \(\displaystyle \,x=0,\:y=0,\) below by \(\displaystyle z=2,\) and above by \(\displaystyle z\:=\:4\,-\,x^2\,-\,y^2\)

I figured my limits for \(\displaystyle x\) would be: \(\displaystyle 0\) to \(\displaystyle 2.\;\;\) no

The intersection of the two surfaces is: \(\displaystyle \:2 \:=\:4\,-\,x^2\,-\,y^2\;\;\Rightarrow\;\;x^2\,+\,y^2\:=\:2\)
. . a circle with radius \(\displaystyle \sqrt{2}\)

The triple integral would be: \(\displaystyle \L\int^{\;\;\;\sqrt{2}}_0\int^{\;\;\;\sqrt{2}}_0\int^{\;\;\;\;\;\;\;\;\;\;\;4-x^2-y^2}_2dz\,dy\,dx\)

This is not that complicated ... if we simplify as we work.


The innermost integral gives us: \(\displaystyle \: z\,\L|\)\(\displaystyle \begin{array}{c}4-x^2-y^2 \\ \\ \\ \\ 2\;\;\;\;\;\;\end{array}\)

. . \(\displaystyle =\;\;(4\,-\,x^2\,-\,y^2)\,-\,2\;\;=\;\;2\,-\,x^2\,-\,y^2\)


The next integral is: \(\displaystyle \L\int^{\;\;\;\sqrt{2}}_0\)\(\displaystyle (2\,-\,x^2\,-\,y^2)\,dy\;\;=\;\;2y\,-\,x^2y\,-\,\frac{y^3}{3}\,\L|\)\(\displaystyle \begin{array}{c}\sqrt{2} \\ \\ \\ 0\end{array}\)

. . Factor: \(\displaystyle \;\frac{y}{3}\left(6\,-\,3x^2\,-\,y^2\right)\,\L|\)\(\displaystyle \begin{array}{c}\sqrt{2}\\ \\ \\ 0\end{array}\;\;=\;\;\frac{\sqrt{2}}{3}\left(6\,-\,3x^2\,-\,2)\;\;=\;\;\frac{\sqrt{2}}{3}\left(4\,-\,3x^2\right)\)


The final integral is: \(\displaystyle \:\frac{\sqrt{2}}{3}\L\int^{\;\;\;\sqrt{2}}_0\)\(\displaystyle (4\,-\,3x^2)\,dx \;\;=\;\;\frac{\sqrt{2}}{3}\left(4x\,-\,x^3\right)\,\L|\)\(\displaystyle \begin{array}{c}\sqrt{2} \\ \\ \\ 0\end{array}\)

. . Factor: \(\displaystyle \;\frac{\sqrt{2}}{3}x\left(4\,-\,x^2\right)\,\L|\)\(\displaystyle \begin{array}{c}\sqrt{2}\\ \\ \\ 0\end{array} \;\;=\;\;\frac{\sqrt{2}}{3}\cdot\sqrt{2}(4\,-\,2)\;\;=\;\;\frac{2}{3}(2)\;\;=\;\;\L\fbox{\frac{4}{3}}\)