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UPSC Maths 2013 Paper 1 Q3c — Solution

15 marks · Section A

Question

Evaluate $\displaystyle\iint_D xy\,dA$, where $D$ is the region bounded by the line $y=x-1$ and the parabola $y^{2}=2x+6$.

Technique

$y$-first iterated integral with the parabola/line boundaries given as $x=$ function of $y$; polynomial expansion and term-by-term evaluation.

Solution

Strategy. Find intersection points; integrate in the natural order (using $y$ as the outer variable since both boundaries are easily expressible as $x=$ function of $y$).

Step 1 — Find intersections

Set $y=x-1\Rightarrow x=y+1$, substitute into $y^{2}=2x+6$:

So $y=-2$ or $y=4$. The corresponding $x$-values: $x=-1$ and $x=5$. The two curves meet at $(-1,-2)$ and $(5,4)$.

Step 2 — Determine left/right boundaries

For $y$ between $-2$ and $4$:

• The parabola gives $x_{\text{parab}}=\dfrac{y^{2}-6}{2}$.
• The line gives $x_{\text{line}}=y+1$.

At $y=0$: parabola gives $x=-3$, line gives $x=1$. So parabola is on the left and line is on the right throughout the strip. (Indeed, $x_{\text{line}}-x_{\text{parab}}=y+1-(y^{2}-6)/2=-(y-4)(y+2)/2$, which is positive for $-2<y<4$.)

Step 3 — Set up the iterated integral

Inner integral: $\int_{(y^{2}-6)/2}^{y+1}xy\,dx=y\left[\dfrac{x^{2}}{2}\right]_{(y^{2}-6)/2}^{y+1}=\dfrac{y}{2}\!\left[(y+1)^{2}-\left(\dfrac{y^{2}-6}{2}\right)^{2}\right]$.

Step 4 — Simplify the bracket

$(y+1)^{2}=y^{2}+2y+1$.

$\left(\dfrac{y^{2}-6}{2}\right)^{2}=\dfrac{y^{4}-12y^{2}+36}{4}$.

Bracket $=\dfrac{4(y^{2}+2y+1)-(y^{4}-12y^{2}+36)}{4}=\dfrac{-y^{4}+16y^{2}+8y-32}{4}$.

Thus

Step 5 — Evaluate the polynomial integral

Term-by-term over $[-2,4]$:

Sum: $-672+960+192-192=288$.

Therefore

Answer