Find the volume of the solid bounded by the coordinate
Find the volume of the solid bounded by the coordinate planes, the plane: $$ 2x-3y-12=0 $$ and the cylinder: $$ z=\frac12y^2. $$
Explanation of Method
A note on the problem statement: the plane exactly as given, $2x-3y-12=0$ (i.e. $x=\frac{3y+12}2$), does not close off a bounded region in the first octant: for every $y\geqslant0$ this gives $x\geqslant6>0$, so $y$ is left completely unrestricted and the region extends to infinity. This is almost certainly a sign transcription slip for $2x+3y-12=0$ (i.e. $x=\frac{12-3y}2$), which meets both axes at finite points ($x=6$ when $y=0$; $y=4$ when $x=0$) and bounds a genuine triangle together with the coordinate axes — exactly the kind of region this type of problem requires. The solution below proceeds with this corrected plane.
- Identify the triangular base region cut out by $x=0,\,y=0,\,2x+3y=12$.
- Set up the volume integral with height $z=\frac12y^2$.
- Evaluate the resulting integral.
Step 1 · Identify the base region
The plane $2x+3y=12$ meets $y=0$ at $x=6$ and $x=0$ at $y=4$, giving the triangle with vertices $(0,0),(6,0),(0,4)$:
$$ D = \left\{(x,y):0\leqslant x\leqslant6,\ 0\leqslant y\leqslant\frac{12-2x}3\right\}. $$
Step 2 · Set up and evaluate the volume integral
$$ V = \int_0^6\!\int_0^{(12-2x)/3}\frac12y^2\,dy\,dx = \int_0^6\frac16\left(\frac{12-2x}3\right)^3dx = \frac1{162}\int_0^6(12-2x)^3\,dx. $$
Substituting $u=12-2x$ ($du=-2\,dx$; $u:12\to0$ as $x:0\to6$):
$$ V = \frac1{162}\cdot\frac12\int_0^{12}u^3\,du = \frac1{324}\left[\frac{u^4}4\right]_0^{12} = \frac{12^4}{1296} = \frac{20736}{1296} = 16. $$
Additional Notes
- A note on the source computation: the source correctly recognized that the plane exactly as stated ($2x-3y-12=0$) leaves the region unbounded in $y$, and worked through several unsuccessful attempts to reconcile this before settling on the same corrected plane $2x+3y=12$ used above, reaching the same final answer.
- Verified symbolically: direct evaluation of the nested integral over the triangle bounded by $2x+3y=12$ gives exactly $16$.
Result
Using the corrected plane $2x+3y=12$, the volume is: