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Calculus 3 : Double Integration with Change of Variables

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Example Question #681 : Multiple Integration

Evaluate $\int \int_R 3x+2y \ dA$ , where is the trapezoidal region with vertices given by (0,0) , (5,0) , $(\frac{5}{2}, \frac{5}{2})$ , and $(\frac{5}{2}, -\frac{5}{2})$ ,

using the transformation x=2u+3v , and y=2u-3v .

Possible Answers:

$\frac{35}{4}$

$\frac{375}{4}$

$\frac{37}{4}$

$\frac{375}{2}$

Correct answer:

$\frac{375}{4}$

Explanation:

The first thing we have to do is figure out the general equations for the lines that create the trapezoid.

y=x

y=-x

y=-x+5

y=x-5

Now we have the general equations for out trapezoid, now we need to plug in our transformations into these equations.

y=x

2u+3v=2u-3v

$-6v=0\rightarrow v=0$

y=-x

2u+3v=-2u+3v

$4u=0\rightarrow u=0$

y=-x+5

2u+3v=-2u+3v+5

$4u=5\rightarrow u=\frac{5}{4}$

y=x-5

2u+3v=2u-3v-5

$-6v=-5\rightarrow v=\frac{5}{6}$

So our region is a rectangle given by $0 \leq u \leq \frac{5}{4}$ , $0 \leq v \leq \frac{5}{6}$

Next we need to calculate the Jacobian.

$\begin{vmatrix} 2 & 3\\ 2 & -3 \end{vmatrix} =-6-6=-12$

Now we can put the integral together.

$\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} (3(2u+3v)+2(2u-3v))) \cdot \left | -12\right | \ du\ dv$

$=\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} 120u+36v \ du\ dv$

$=\int_{0}^{\frac{5}{6}} 60u^2+36uv \Big|_{0}^{\frac{5}{4}}\ dv$

$=\int_{0}^{\frac{5}{6}} 60( \frac{5}{4})^2+36(\frac{5}{4})v-0 \ dv$

$=\int_{0}^{\frac{5}{6}} \frac{375}{4}+45v \ dv$

$=\frac{375}{4}v+\frac{45}{2}v^2\Big|_{0}^{\frac{5}{6}} \ dv$

$=\frac{375}{4}\cdot\frac{5}{6}+\frac{45}{2}\cdot\Big(\frac{5}{6}\Big)^2 -0=\frac{375}{4}$

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Calculus 3 : Double Integration with Change of Variables

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Example Question #681 : Multiple Integration

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