← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q2c — Step-by-Step Solution

15 marks · Section A

Jacobian · Calculus · asked 4× in 14 yrs · Read the full method →

Question

By using the transformation $x+y=u,\;y=uv$, evaluate the integral $\displaystyle\iint\{xy(1-x-y)\}^{1/2}\,dx\,dy$ taken over the area enclosed by the straight lines $x=0,\;y=0$ and $x+y=1$.

Technique

$(u,v)$ substitution turns the triangle into a unit square; integrand separates; recognise standard Beta-function integrals.

Solution

Strategy. The given substitution maps the triangle $\{x,y\ge 0,\,x+y\le 1\}$ to the unit square $\{0\le u,v\le 1\}$. After substitution, the integral separates into two Beta-function integrals.

Step 1 — Inverse substitution and Jacobian

$u=x+y,\;v=y/(x+y)$, so $y=uv,\;x=u(1-v)$.

Boundaries:

• $x=0\Rightarrow u(1-v)=0\Rightarrow v=1$ (for $u\ne 0$).
• $y=0\Rightarrow uv=0\Rightarrow v=0$.
• $x+y=1\Rightarrow u=1$.

Image in $(u,v)$ space: unit square $0\le u\le 1$, $0\le v\le 1$.

Jacobian:

So $dx\,dy=u\,du\,dv$.

Step 2 — Transform the integrand

$xy=u(1-v)\cdot uv=u^{2}v(1-v)$. $1-x-y=1-u$. $xy(1-x-y)=u^{2}v(1-v)(1-u)$. $\sqrt{xy(1-x-y)}=u\sqrt{v(1-v)(1-u)}$.

Step 3 — Set up and separate

Step 4 — Evaluate each Beta integral

Integral 1: $\displaystyle\int_0^{1}u^{2}(1-u)^{1/2}\,du=B(3,3/2)=\dfrac{\Gamma(3)\Gamma(3/2)}{\Gamma(9/2)}$.

$\Gamma(3)=2!=2$, $\Gamma(3/2)=\dfrac{\sqrt\pi}{2}$, $\Gamma(9/2)=\dfrac{7}{2}\cdot\dfrac{5}{2}\cdot\dfrac{3}{2}\cdot\dfrac{1}{2}\sqrt\pi=\dfrac{105\sqrt\pi}{16}$.

Integral 2: $\displaystyle\int_0^{1}v^{1/2}(1-v)^{1/2}\,dv=B(3/2,3/2)=\dfrac{\Gamma(3/2)^{2}}{\Gamma(3)}=\dfrac{\pi/4}{2}=\dfrac{\pi}{8}$.

Step 5 — Multiply

Answer