# Use the change of variables s=xy and t = xy^2 to compute the integral of xy^2 with respect to dA...

## Question:

Use the change of variables {eq}s=xy {/eq} and {eq}t = xy^2 {/eq} to compute the integral {eq}\int_R xy^2 \mathrm{d}A {/eq}, where {eq}R {/eq} is the region bounded by {eq}xy = 1 {/eq}, {eq}xy = 6 {/eq}, {eq}xy^2=1 {/eq}, and {eq}xy^2=6 {/eq}.

## Calculation of Double Integration Using Transformation:

Let us suppose we want to evaluate: {eq}\iint_{D} f(x,y) dx dy {/eq}, which can be evaluated by transforming the variables {eq}{\rm{x = }}{{\rm{f}}_1}(u,v) {/eq} and {eq}{\rm{y = }}{{\rm{f}}_2}(u,v) {/eq} and the evaluation of integration on the U-V plane after transformation is given by:

{eq}I=\iint_{D} f(x,y) dx dy=\iint_{D} f(u,v)|J| du dv {/eq} ,where {eq}{\rm{J = }}\frac{{\partial (x,y)}}{{\partial (u,v)}} = \left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}} & {\frac{{\partial x}}{{\partial v}}} \\ {\frac{{\partial y}}{{\partial u}}} & {\frac{{\partial y}}{{\partial v}}} \\ \end{array}} \right| {/eq}

## Answer and Explanation:

Here we have to evaluate: {eq}\int_R xy^2 \mathrm{d}A {/eq} where {eq}R {/eq} is the region bounded by {eq}xy = 1 {/eq}, {eq}xy = 6 {/eq}, {eq}xy^2=1 {/eq}, and {eq}xy^2=6 {/eq}.

Let us choose a transformation {eq}s = xy {/eq} and {eq}t=xy^2 {/eq}

Hence:

{eq}{\rm{J = }}\frac{{\partial (x,y)}}{{\partial (s,t)}} = \frac{1}{{\left( {\frac{{\partial (s,t)}}{{\partial (x,y)}}} \right)}} = \frac{1}{{\left| {\begin{array}{*{20}{c}} {\frac{{\partial s}}{{\partial x}}} & {\frac{{\partial s}}{{\partial y}}} \\ {\frac{{\partial t}}{{\partial x}}} & {\frac{{\partial t}}{{\partial y}}} \\ \end{array}} \right|}} = \frac{1}{{\left| {\begin{array}{*{20}{c}} y & x \\ {{y^2}} & {2xy} \\ \end{array}} \right|}} = \frac{1}{{x{y^2}}} = \frac{1}{t} {/eq}

And the updated limits can be given as: {eq}s:1 \to 6 {/eq} and {eq}t:1 \to 6 {/eq}

So, after transformation, the evaluation of the integration is given by:

{eq}{\rm{I = }}\int\limits_{t = 1}^6 {\int\limits_{s = 1}^6 {\left( {x{y^2}} \right)\left| J \right|dsdt} } \\ = \int\limits_{t = 1}^6 {\int\limits_{s = 1}^6 {\left( t \right)\left| {\frac{1}{t}} \right|dsdt} } \\ = \int\limits_{t = 1}^6 {\int\limits_{s = 1}^6 {dsdt} } \\ = 25 \\ {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 15