# Use the transformation x = u + v, y = y = u - v to evaluate the double integral over R of (x +...

## Question:

Use the transformation {eq}\; x = u + v, \; y = y = u - v \; {/eq} to evaluate {eq}\iint_{R} (x + 2y) \, \mathrm{d}A_{x, y} {/eq} where {eq}R {/eq} is the square region in the {eq}xy {/eq}-plane with vertices {eq}\; (0, 0), \; (1, 1), \; (2, 0), {/eq} and {eq}\; (1, -1) {/eq}.

## Using Transformation On a Parallelogram:

When given a problem where we have to integrate a scalar function over a parallelogram we should almost always consider using substitution.

A parallelogram is nothing more than a transformed rectangle.

When we find a transformation {eq}L {/eq} such that {eq}L(R) {/eq} is the given parallelogram we can apply substitution and integrate over a rectangle easily by using Fubini's theorem:

{eq}\int_{L(R)} f(x,y) \,dx \,dy= \int_{R} f(L(u,v)) | J_{L} | \,du \,dv {/eq}

## Answer and Explanation:

We are already given the transformation in the problem itself:

{eq}L(u,v) = (u+v, u-v) = (x,y) {/eq}

Which gives us {eq}|J_L |= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = |-2| = 2 {/eq}

Now we have to find the rectangle that transforms into this parallelogram.

First we solve the given two equations for u and v:

{eq}u = \frac{x+y}{2} \\ v= \frac{x-y}{2} {/eq}

The given parallelogram is defined by the lines:

{eq}x-y = 0 \\ x-y= 2 \\ x+y = 0 \\ x+y = 2 {/eq}

from which we conclude {eq}u \in [0,1] , v\in [0,1]. {/eq}

Now we have everythin we need to find the integral:

{eq}\begin{align*} I&= \int_R (x+2y) \,dA \\ &= \int_0^1 \int_0^1 (u+v + 2u - 2v )\cdot 2 \,du \,dv \\ &= 2\int_0^1 \int_0^1 (3u - v) \,du \,dv \\ &= 2 \int_0^1 \frac{3}{2} - v \,dv \\ &= 2 (\frac{3}{2} - \frac{1}{2}) \\ &= 2 \end{align*} {/eq}

NOTE;

The first area we are given is a square but I used the term parallelogram because it is more general. Almost the exact approach could've been used in any case where we are given a parallelogram.

#### Learn more about this topic: Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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