Let I=\iint_D x^2y^2 dxdy, where ={(x,y):1\leq xy\leq 3,0\leq x^y\leq 7,x\geq 0,y\geq 0} Show...


Let {eq}I=\iint_D x^2y^2 dxdy{/eq}, where {eq}D={(x,y):1\leq xy\leq 3,0\leq x^y\leq 7,x\geq 0,y\geq 0} {/eq} Show that the mapping {eq}u=xy, v=x^y{/eq} maps {eq}D{/eq} to the rectangle {eq}R=[1,3]\times [0,7]. {/eq}

(a) Compute {eq}\frac{\partial(x,y)}{\partial(u,v)}{/eq} by first computing {eq}\frac{\partial(u,v)}{\partial(x,y)} {/eq}

(b) Use the Change of Variables Formula to show that {eq}I{/eq} is equal to the integral of {eq}f(u,v)=v{/eq} over {eq}R{/eq} and evaluate.

Change of variables in double integrals:

When we evaluate a multiple integral, the integration region may be complex. A change of variable can allow us to map the region of integration to a space in which the region result less complex. However, when we make the change of variables we have to take into account that the differential of area or volume changes. The Jacobian, {eq}J {/eq}, or determinant of the Jacobian matrix, allows us to map this differential. For example

{eq}dxdy=|J|dudv {/eq}

where {eq}J=\frac{\delta(x, y)}{\delta(u, v)} {/eq} is the determinant of the Jacobian Matrix. This determinant satisfies that

{eq}\frac{\delta(x, y)}{\delta(u, v)}\times \frac{\delta(u, v)}{\delta(x, y)}=1 {/eq}

Answer and Explanation:

It is given the region

{eq}D=\{(x, y) | 1 \leq x y \leq 5, \quad 0 \leq x-y \leq 7, \quad x \geq 0, \quad y \geq 0\} {/eq}

If we use the transformations

{eq}u=x y, \quad v=x-y {/eq}

we have that

{eq}\begin{array}{l} 1 \leq x y \leq 5\quad \Rightarrow\quad 1 \leq u \leq 5 \\ 0 \leq x-y \leq 7 \quad \Rightarrow \quad 0 \leq v \leq 7 \end{array} {/eq}

Therefore, this transformations maps the given region to a rectangle in {eq}u,v {/eq} coordinates

{eq}R=[1,5] \times[0,7] {/eq}

Now, we compute the Jacobian

{eq}\frac{\delta(u, v)}{\delta(x, y)}=\left|\begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array}\right|=\left|\begin{array}{cc} y & x \\ 1 & -1 \end{array}\right|=-y-x {/eq}

Taking into account that

{eq}\frac{\delta(u, v)}{\partial(x, y)} \times\frac{\delta(x, y)}{\delta(u, v)}=1 {/eq}

we have that

{eq}\frac{\delta(x, y)}{\delta(u, v)}=-\frac{1}{x+y} {/eq}

Finally, the given integral can evaluated

{eq}\displaystyle I=\iint_{D}\left(x^{2}-y^{2}\right) d x d y=\iint_{R} v d u d v=\int_{1}^{5} d u \int_{0}^{7} v d v=98 {/eq}

where we have used that

{eq}\begin{aligned} \left(x^{2}-y^{2}\right) d x d y &=(x+y)(x-y) d x d y \\ &=v(x+y)\left|\frac{\delta(x, y)}{\delta(u, v)}\right|dudv \\ &=v(x+y) \frac{1}{x+y} d u d v=v d u d v \end{aligned} {/eq}

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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