Question

(For those who have studied multivariable calculus.) Let Tbe an invertible linear transformation from${\mathbf{ℝ}}^{\mathbf{2}}$to${\mathbf{ℝ}}^{\mathbf{2}}$, represented by the matrix M. Let${\mathit{\Omega }}_{\mathbf{1}}$be the unit square in ${\mathbf{ℝ}}^{\mathbf{2}}$and${\mathit{\Omega }}_{\mathbf{2}}$its image under T . Consider a continuous function$\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$from${\mathbf{ℝ}}^{\mathbf{2}}$to$\mathbf{ℝ}$, and define the function$\mathbf{g}\mathbf{(}\mathbf{u}\mathbf{,}\mathbf{v}\mathbf{)}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{T}\mathbf{\right(}\mathbf{u}\mathbf{,}\mathbf{v}\mathbf{\left)}\mathbf{\right)}$. What is the relationship between the following two double integrals?

${\mathbf{\iint }}_{\mathbf{\Omega }\mathbf{2}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dA}$and${\mathbf{\iint }}_{\mathbf{\Omega }\mathbf{1}}\mathbf{g}\mathbf{(}\mathbf{u}\mathbf{,}\mathbf{v}\mathbf{)}\mathbf{dA}$

Your answer will involve the matrix M. Hint: What happens when$\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{1}$, for all$\mathit{x}\mathbf{,}\mathit{y}$?

Short answer

Therefore,the relationship between the given two double integralsis given by,

${\iint }_{{\Omega }_2}f(x,y)dA=|detM|{\iint }_{{\Omega }_2}g(u,v)dA$, and the given two double integrals are same.

Step-by-step solution

Matrix Definition. 

Matrix is a set of numbers arranged in rows and columns so as to form a rectangulararray.

The numbers are called the elements, or entries, of the matrix.

If there are rows and columns, the matrix is said to be an “ mby n” matrix, written “$m\times n$.”

What is the relationship between the given two double integrals.

For,$f(x,y)=1,\forall (x,y)\in {\mathrm{ℝ}}^2,$,

We have

${\iint }_{{\Omega }_2}f(x,y)dA={\iint }_{{\Omega }_2}1dA=|detM|$

Which is the area of${\Omega }_2$.

On the other hand,

${\iint }_{{\Omega }_1}g(u,v)dA={\iint }_{{\Omega }_1}f\left(T\right(u,v\left)\right)dA={\iint }_{{\Omega }_1}1dA=1$

Which is the area of.

From multivariable calculus, we know that, for a differentiable injective function $\phi :U\to {\mathrm{ℝ}}^n$where$U\subseteq {\mathrm{ℝ}}^n$ is an open set, and D is the differentiation matrix of $\phi$, and a continuous function $f:\phi \left(U\right)\to \mathrm{ℝ}$, applies,

${\int }_{\phi \left(U\right)}f\left(x\right)dx={\int }_{U}f\left(\phi \right(u\left)\right)|detD(u\left)\right|du$.

In this case, is an invertible linear transformation, which means it's a differentiable injective function, thus we can apply the upper theorem here, giving us

$$\begin{gathered} {\iint }_{{}_{(}{\Omega }_2}f(x,y)dA={\iint }_{T_T\left({\Omega }_1\right)}f(x,y)dA \\ {\iint }_{{\Omega }_2}f(x,y)dA={\iint }_{{\Omega }_1}f\left(T\right(u,v\left)\right)|detM|dA \\ {\iint }_{{\Omega }_2}f(x,y)dA=|detM|{\iint }_{{\Omega }_2}g(u,v)dA \end{gathered}$$

Therefore,

${\iint }_{{\Omega }_2}f(x,y)dA=|detM|{\iint }_{{\Omega }_2}g(u,v)dA$.