Question

The value of  ∬R (x-y)^2 dx dy where R is the parallelogram with vertices (0,0), (1,1),(2,0), (1,-1) when solved using change of variables is given by____

(a) 16/3

(b) 8/3

(c) 4/3

(d) 0

The question was asked in examination.

My question is based upon Change of Variables In a Double Integral topic in section Multiple Integrals of Engineering Mathematics

Answer 1

Right answer is (b) 8/3

To explain I would say: W.K.T from change of variables principle

\(\int\int_R f(x,y)\,dx \,dy = \int\int_S f(g(u,v),h(u,v)) \frac{∂(x,y)}{∂(u,v)} \,du \,dv …..(1)\)

From the above diagram in the region R the equations are given by

x-y=0, x-y=2, x+y=0, x+y=2 from this we can observe that change of

variables is u=x-y, v=x+y solving we get \(x=\frac{u+v}{2}, y=\frac{v-u}{2}\)

\(\frac{∂(x,y)}{∂(u,v)} = \begin{vmatrix}

\frac{∂x}{∂u}&\frac{∂x}{∂v}\\

\frac{∂y}{∂u}&\frac{∂y}{∂v}\\

\end{vmatrix} = \begin{vmatrix}

0.5 &0.5\\

-0.5&0.5\\

\end{vmatrix} = 0.5\)

The region S in the (u,v) is the square 0<u<2,0<v<2. Since x-y=u integral

becomes \(\int_0^2 \int_0^2 0.5u^2 \,du \,dv\)……from(1)

\(=\int_0^2 \Big[\frac{u^3}{6}\Big]_0^2 \,dv  = \int_0^2 \frac{4}{3} dv = \frac{8}{3}.\)