Calculus III - Triple Integrals

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Section 15.5 : Triple Integrals

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1. Evaluate $\displaystyle \int_{2}^{3}{{\int_{{ - 1}}^{4}{{\int_{1}^{0}{{4{x^2}y - {z^3}\,dz}}\,dy}}\,dx}}$

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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.

So, here is the $z$ integration.

$\begin{align*}\int_{2}^{3}{{\int_{{ - 1}}^{4}{{\int_{1}^{0}{{4{x^2}y - {z^3}\,dz}}\,dy}}\,dx}} & = \int_{2}^{3}{{\int_{{ - 1}}^{4}{{\left. {\left( {4{x^2}yz - \frac{1}{4}{z^4}} \right)} \right|_1^0\,dy}}\,dx}}\\ & = \int_{2}^{3}{{\int_{{ - 1}}^{4}{{\frac{1}{4} - 4{x^2}y\,dy}}\,dx}}\end{align*}$

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 $z$ integration the $x$ ’s and $y$ ’s are all considered to be constants.

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Next, we’ll do the $y$ integration.

$\begin{align*}\int_{2}^{3}{{\int_{{ - 1}}^{4}{{\int_{1}^{0}{{4{x^2}y - {z^3}\,dz}}\,dy}}\,dx}} & = \int_{2}^{3}{{\left. {\left( {\frac{1}{4}y - 2{x^2}{y^2}} \right)} \right|_{ - 1}^4\,dx}}\\ & = \int_{2}^{3}{{\frac{5}{4} - 30{x^2}\,dx}}\end{align*}$ Show Step 3

Finally, we’ll do the $x$ integration.

$\int_{2}^{3}{{\int_{{ - 1}}^{4}{{\int_{1}^{0}{{4{x^2}y - {z^3}\,dz}}\,dy}}\,dx}} = \left. {\left( {\frac{5}{4}x - 10{x^3}} \right)} \right|_2^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{755}}{4}}}$

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