Paul's Online Notes
Paul's Online Notes
Home / Calculus III / Multiple Integrals / Triple Integrals
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 15.5 : Triple Integrals

2. Evaluate \( \displaystyle \int_{0}^{1}{{\int_{0}^{{{z^{\,2}}}}{{\int_{0}^{3}{{y\cos \left( {{z^5}} \right)\,dx}}\,dy}}\,dz}}\)

Show All Steps Hide All Steps

Start Solution

There really isn’t all that much to this problem. All we need to do is integrate following the given order and recall that just like with double integrals we start with the “inside” integral and work our way out.

Also note that the fact that one of the limits is not a constant is not a problem. There is nothing that says that triple integrals set up as this is must only have constants as limits!

So, here is the \(x\) integration.

\[\begin{align*}\int_{0}^{1}{{\int_{0}^{{{z^{\,2}}}}{{\int_{0}^{3}{{y\cos \left( {{z^5}} \right)\,dx}}\,dy}}\,dz}} & = \int_{0}^{1}{{\int_{0}^{{{z^{\,2}}}}{{\left. {\left( {y\cos \left( {{z^5}} \right)\,x} \right)} \right|_0^3\,dy}}\,dz}}\\ & = \int_{0}^{1}{{\int_{0}^{{{z^{\,2}}}}{{3y\cos \left( {{z^5}} \right)\,\,dy}}\,dz}}\end{align*}\]

Remember that triple integration is just like double integration and all the variables other than the one we are integrating with respect to are considered to be constants. So, for the \(x\) integration the \(y\)’s and \(z\)’s are all considered to be constants.

Show Step 2

Next, we’ll do the \(y\) integration.

\[\begin{align*}\int_{0}^{1}{{\int_{0}^{{{z^{\,2}}}}{{\int_{0}^{3}{{y\cos \left( {{z^5}} \right)\,dx}}\,dy}}\,dz}}& = \int_{0}^{1}{{\left. {\left( {\frac{3}{2}{y^2}\cos \left( {{z^5}} \right)} \right)} \right|_0^{{z^2}}\,dz}}\\ & = \int_{0}^{1}{{\frac{3}{2}{z^4}\cos \left( {{z^5}} \right)\,dz}}\end{align*}\] Show Step 3

Finally, we’ll do the \(z\) integration and note that the only way we are able to do this integration is because of the \({z^4}\)that is now in the integrand. Without that present we would not be able to do this integral.

\[\int_{0}^{1}{{\int_{0}^{{{z^{\,2}}}}{{\int_{0}^{3}{{y\cos \left( {{z^5}} \right)\,dx}}\,dy}}\,dz}} = \left. {\left( {\frac{3}{{10}}\sin \left( {{z^5}} \right)} \right)} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{{10}}\sin \left( 1 \right) = 0.2524}}\]

So, not too much to do with this problem since the limits were already set up for us.