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### Section 4-3 : Double Integrals over General Regions

10. Evaluate $$\displaystyle \int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}}$$ by first reversing the order of integration.

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Let’s start off by noticing that if we were to integrate with respect to $$y$$ first we’d need to do a trig substitution (which we’ll all be thankful if we don’t need to do it!) so interchanging the order in this case might well save us a messy integral.

So, here are the limits we get from the integral.

$\begin{array}{c}0 \le x \le 3\\ 2x \le y \le 6\end{array}$

Here is a quick sketch of the region these limits describe. When reversing the order of integration it is often very helpful to have a sketch of the region to make sure we get the correct limits for the reversed order.

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Okay, if to reverse the order of integration we need to integrate with respect to $$x$$ first. The limits for the reversed order are then,

$\begin{array}{c}0 \le y \le 6\\ 0 \le x \displaystyle \le \frac{1}{2}y\end{array}$

The integral with reversed order is,

$\int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}} = \int_{0}^{6}{{\int_{0}^{{\frac{1}{2}y}}{{\sqrt {{y^2} + 2} \,dx}}\,dy}}$ Show Step 3

Now all we need to do is evaluate the integrals. Here is the $$x$$ integration.

$\int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}} = \int_{0}^{6}{{\left. {\left( {x\sqrt {{y^2} + 2} } \right)} \right|_0^{\frac{1}{2}y}\,dy}} = \int_{0}^{6}{{\frac{1}{2}y\sqrt {{y^2} + 2} \,dy}}$ Show Step 4

Note that because of the $$x$$ integration we’ll not need to do a trig substitution. All we need is a simple Calculus I integral. Here is the $$y$$ integration (we’ll leave it to you to verify the substitution details).

$\int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}} = \left. {\left( {\frac{1}{6}{{\left( {{y^2} + 2} \right)}^{\frac{3}{2}}}} \right)} \right|_0^6 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{6}\left( {{{38}^{\frac{3}{2}}} - {2^{\frac{3}{2}}}} \right) = 38.5699}}$