Find the volume of the solid bounded by the cylinders
Find the volume of the solid bounded by the cylinders: $$ y=\sqrt x,\qquad y=2\sqrt x $$ and the planes: $$ z=0,\qquad x+z=6. $$
Explanation of Method
The base region lies between the two parabolic cylinders $y=\sqrt x$ and $y=2\sqrt x$, and the height is $z=6-x$ (from $x+z=6$), which stays non-negative for $x\leqslant6$.
- Identify the base region $D=\{(x,y):0\leqslant x\leqslant6,\ \sqrt x\leqslant y\leqslant2\sqrt x\}$.
- Set up the volume integral with height $z=6-x$.
- Evaluate the resulting integral.
Step 1 · Identify the region
Since $z=6-x\geqslant0$ requires $x\leqslant6$, and $x\geqslant0$ for the square roots to be defined, the base region is $0\leqslant x\leqslant6,\ \sqrt x\leqslant y\leqslant2\sqrt x$.
Step 2 · Set up and evaluate the volume integral
$$ V = \int_0^6\!\int_{\sqrt x}^{2\sqrt x}(6-x)\,dy\,dx = \int_0^6(6-x)\big(2\sqrt x-\sqrt x\big)\,dx = \int_0^6(6-x)\sqrt x\,dx. $$
$$ = \int_0^6\big(6x^{1/2}-x^{3/2}\big)\,dx = 6\left[\frac{2x^{3/2}}3\right]_0^6-\left[\frac{2x^{5/2}}5\right]_0^6 = 4\cdot6^{3/2}-\frac{2\cdot6^{5/2}}5. $$
Using $6^{3/2}=6\sqrt6$ and $6^{5/2}=36\sqrt6$:
$$ V = 24\sqrt6-\frac{72\sqrt6}5 = \sqrt6\left(\frac{120-72}5\right) = \frac{48\sqrt6}5. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: direct evaluation of $\int_0^6(6-x)\sqrt x\,dx$ gives exactly $\dfrac{48\sqrt6}5\approx23.5151$.
Result
The volume is: