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UPSC 2026 Maths Optional Paper 1 Q3b-ii — Step-by-Step Solution

10 marks · Section A

Areas, surface areas, volumes via integration · Calculus · asked 5× in 14 yrs · Read the full method →

Question

Find the volume of the solid, which is formed from the intersection of the coordinate planes $x = 0$, $y = 0$, $z = 0$ and the plane $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$.

Technique

Set up $V=\iiint_{E}dV$ over the tetrahedron and integrate in the order $dz\,dy\,dx$: project onto the $xy$-plane to get the triangular shadow, take $z$ from the floor $z=0$ up to the roof $z=c\left(1-\frac xa-\frac yb\right)$, and let the limits cascade. The marks are for the correct cascade of limits, not for quoting $\frac{abc}{6}$.

Solution

Two words in the question need fixing before any integration. Read “the intersection of the coordinate planes … and the plane” as “the solid bounded by them”: the set-theoretic intersection of four planes is empty (or a point or a line) and has zero volume, so the intended region is the first-octant tetrahedron identified in Step 1. And throughout take $a,b,c>0$ — the paper does not state it, but it is what makes those four planes bound a genuine solid in the first octant (for arbitrary non-zero signs the volume is $\frac{|abc|}{6}$; see Step 7).

Step 1 — Identify the solid.

The plane $\frac xa+\frac yb+\frac zc=1$ meets the axes at $A(a,0,0)$, $B(0,b,0)$, $C(0,0,c)$; together with the origin $O(0,0,0)$ these are the four vertices. The solid is the tetrahedron $OABC$:

and $V=\displaystyle\iiint_{E}dx\,dy\,dz$.

Step 2 — Limits for $z$ (innermost).

For a fixed $(x,y)$ in the shadow, $z$ runs from the floor $z=0$ up to the slanted roof obtained by solving the plane equation for $z$:

So $0\le z\le c\left(1-\frac xa-\frac yb\right)$.

Step 3 — Limits for the shadow region in the $xy$-plane.

The roof is at or above the floor precisely when $1-\frac xa-\frac yb\ge 0$. Hence the projection of $E$ onto $z=0$ is the triangle

the triangle $OAB$ with hypotenuse the line $\frac xa+\frac yb=1$.

The full iterated integral is therefore

Step 4 — Integrate in $z$.

Step 5 — Integrate in $y$.

Write $u = 1-\dfrac xa$ (a constant for the $y$-integration), so the $y$-limit is $y=bu$ and the integrand is $c\left(u-\frac yb\right)$:

That is,

Step 6 — Integrate in $x$.

Since $\dfrac{d}{dx}\left(1-\dfrac xa\right)^{3}=-\dfrac{3}{a}\left(1-\dfrac xa\right)^{2}$,

Hence

Step 7 — Consistency check against the elementary formula.

The tetrahedron has base the right triangle $OAB$ in $z=0$, of area $\frac12\,(OA)(OB)=\frac12 ab$, and apex $C(0,0,c)$ at perpendicular height $c$ above that base. The pyramid formula gives

in agreement with the integration. (For a general sign convention, if $a,b,c$ are allowed any non-zero signs the volume is $\frac{|abc|}{6}$.)

Answer