Find the volume of the solid bounded by the coordinate
Find the volume of the solid bounded by the coordinate planes and the planes: $$ x=2,\qquad y=3,\qquad x+y+z=4. $$
Explanation of Method
The solid sits over the rectangle $0\leqslant x\leqslant2,\,0\leqslant y\leqslant3$, with $z$ running from $0$ up to $z=4-x-y$ wherever that is positive. Since $4-x-y$ becomes negative near the far corner $(2,3)$ (where $4-2-3=-1<0$), the rectangle must be split at the line $x+y=4$, and only the sub-region where $x+y\leqslant4$ contributes.
- Check where $z=4-x-y$ stays positive within the rectangle $[0,2]\times[0,3]$.
- Split the rectangle at $x+y=4$ and integrate $z=4-x-y$ over the valid sub-region.
- Evaluate the resulting double integral.
Step 1 · Locate where the ceiling meets the rectangle's far corner
Checking the corners of $[0,2]\times[0,3]$: at $(0,0)$, $z=4>0$; at $(2,0)$, $z=2>0$; at $(0,3)$, $z=1>0$; but at $(2,3)$, $z=-1<0$. So the line $x+y=4$ cuts across the rectangle, crossing $x=2$ at $y=2$ and $y=3$ at $x=1$.
Step 2 · Set up the split integral
For $0\leqslant x\leqslant1$, the full range $y\in[0,3]$ keeps $x+y\leqslant4$; for $1\leqslant x\leqslant2$, only $y\in[0,4-x]$ does. So:
$$ V = \int_0^1\!\int_0^3(4-x-y)\,dy\,dx \ +\ \int_1^2\!\int_0^{4-x}(4-x-y)\,dy\,dx. $$
Step 3 · Evaluate the first piece
$$ \int_0^3(4-x-y)\,dy = \left[(4-x)y-\frac{y^2}2\right]_0^3 = 3(4-x)-\frac92 = \frac{15}2-3x. $$
$$ \int_0^1\left(\frac{15}2-3x\right)dx = \left[\frac{15x}2-\frac{3x^2}2\right]_0^1 = \frac{15}2-\frac32 = 6. $$
Step 4 · Evaluate the second piece
$$ \int_0^{4-x}(4-x-y)\,dy = \frac{(4-x)^2}2. $$
$$ \int_1^2\frac{(4-x)^2}2\,dx = \frac12\left[-\frac{(4-x)^3}3\right]_1^2 = \frac16\left[3^3-2^3\right] = \frac{19}6. $$
Step 5 · Total volume
$$ V = 6+\frac{19}6 = \frac{55}6. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: both pieces evaluate to exactly $6$ and $\frac{19}6$ respectively, giving $\frac{55}6$.
Result
The volume is: