Find the volume of the solid bounded by the plane: frac
Find the volume of the solid bounded by the plane: $$ \frac xa+\frac yb+\frac zc=1 $$ and the coordinate planes.
Explanation of Method
This is the classical intercept-form tetrahedron, with vertices at the origin and at $(a,0,0)$, $(0,b,0)$, $(0,0,c)$. Solving the plane equation for $z$ and integrating over the triangular projection onto the $xy$-plane gives the standard result.
- Solve for $z$ and identify the triangular base region.
- Set up the volume integral.
- Evaluate using the substitution $u=1-x/a$.
Step 1 · Solve for $z$ and identify the base region
$$ z = c\left(1-\frac xa-\frac yb\right). $$
The projection onto the $xy$-plane where $z\geqslant0$, combined with $x,y\geqslant0$, is the triangle with vertices $(0,0)$, $(a,0)$, $(0,b)$.
Step 2 · Set up the volume integral
$$ V = \int_0^a\!\int_0^{b(1-x/a)} c\left(1-\frac xa-\frac yb\right)dy\,dx. $$
Step 3 · Evaluate the inner integral
$$ \int_0^{b(1-x/a)}c\left(1-\frac xa-\frac yb\right)dy = c\left[\left(1-\frac xa\right)y-\frac{y^2}{2b}\right]_0^{b(1-x/a)} = \frac{bc}2\left(1-\frac xa\right)^2. $$
Step 4 · Evaluate the outer integral
Substituting $u=1-\dfrac xa$ ($dx=-a\,du$):
$$ V = \int_0^a\frac{bc}2\left(1-\frac xa\right)^2dx = \frac{abc}2\int_0^1u^2\,du = \frac{abc}2\cdot\frac13 = \frac{abc}6. $$
Additional Notes
- This derivation matches the source exactly. It is the classical formula for the volume of a tetrahedron with legs $a,b,c$ along the three coordinate axes — one-sixth the product of the intercepts.
- Verified symbolically: direct evaluation of the nested integral gives exactly $\dfrac{abc}6$.
Result
The volume is: