Double Integral 147
Medium Polar Coordinates and Change of Variables

Find the volume of the solid bounded by the planes: y=0

Find the volume of the solid bounded by the planes: $$ y=0,\quad z=0,\quad 3x+y=6,\quad 3x+2y=12,\quad x+y+z=6. $$

Explanation of Method

The base region in the $xy$-plane is a triangle cut out by $y=0$, $3x+y=6$, and $3x+2y=12$; the height of the solid over this triangle is $z=6-x-y$ from the plane $x+y+z=6$. Since the triangle's right edge switches from one bounding line to the other partway through, the integral must be split at that point.

  1. Find the vertices of the triangular base by intersecting the three bounding lines.
  2. Set up the volume integral, split where the upper $y$-bound switches from $6-3x$ to $6-\frac32x$.
  3. Evaluate both pieces and sum.

Step 1 · Find the triangle's vertices

Intersecting $y=0$ with $3x+y=6$ gives $(2,0)$; intersecting $y=0$ with $3x+2y=12$ gives $(4,0)$; intersecting $3x+y=6$ with $3x+2y=12$ gives (subtracting) $y=6,\,x=0$, i.e. $(0,6)$. So the base triangle has vertices $(0,6)$, $(2,0)$, $(4,0)$.

Step 2 · Set up the split integral

For $0\leqslant x\leqslant2$, $y$ runs from $0$ to $6-3x$ (bounded by $3x+y=6$); for $2\leqslant x\leqslant4$, $y$ runs from $0$ to $6-\frac32x$ (bounded by $3x+2y=12$):

$$ V = \int_0^2\!\int_0^{6-3x}(6-x-y)\,dy\,dx \ +\ \int_2^4\!\int_0^{6-3x/2}(6-x-y)\,dy\,dx. $$

Step 3 · Evaluate the first piece

$$ \int_0^{6-3x}(6-x-y)\,dy = \frac{(6-3x)(6+x)}2 = \frac{3(2-x)(6+x)}2. $$

$$ \int_0^2\frac{3(2-x)(6+x)}2\,dx = \frac32\int_0^2(12-4x-x^2)\,dx = \frac32\left[12x-2x^2-\frac{x^3}3\right]_0^2 = \frac32\cdot\frac{40}3 = 20. $$

Step 4 · Evaluate the second piece

$$ \int_0^{6-3x/2}(6-x-y)\,dy = \frac{(6-\frac32x)(6-\frac x2)}2 = \frac{3(2-\frac x2)(6-\frac x2)}2. $$

$$ \int_2^4\frac32\left(2-\frac x2\right)\left(6-\frac x2\right)dx = \frac32\int_2^4\left(12-4x+\frac{x^2}4\right)dx = \frac32\left[12x-2x^2+\frac{x^3}{12}\right]_2^4 = \frac32\cdot\frac{14}3 = 7. $$

Step 5 · Total volume

$$ V = 20+7 = 27. $$

Additional Notes

Result

The volume is:

$$ V = 27. $$