Find the volume of the solid bounded by the parabolic
Find the volume of the solid bounded by the parabolic cylinders: $$ y^2=x,\qquad y^2=4x, $$ and the planes: $$ z=0,\qquad x+z=4, $$ with $y>0$.
Explanation of Method
Slicing the solid perpendicular to the $y$-axis is the most direct route: for each fixed $y$, the cylinders pin down a strip in $x$, and the planes $z=0$ and $z=4-x$ pin down the height. Integrating the resulting cross-sectional area over $y$ gives the volume.
- Identify the $x$-strip between the two parabolic cylinders at fixed $y$.
- Compute the cross-sectional area $A(y)=\int(4-x)\,dx$ over that strip.
- Integrate $A(y)$ over the range of $y$.
Step 1 · The $x$-strip at fixed $y$
For $y>0$: $y^2=4x\Rightarrow x=\dfrac{y^2}4$, and $y^2=x\Rightarrow x=y^2$. Since $\dfrac{y^2}4\leqslant y^2$, the strip at height $y$ runs $\dfrac{y^2}4\leqslant x\leqslant y^2$. The solid closes off where the outer boundary $x=y^2$ meets the plane's base $x=4$ (where $z=4-x=0$), i.e. at $y=2$. So $y\in[0,2]$.
Step 2 · Cross-sectional area
$$ A(y)=\int_{y^2/4}^{y^2}(4-x)\,dx = \left[4x-\frac{x^2}2\right]_{y^2/4}^{y^2} = \left(4y^2-\frac{y^4}2\right)-\left(y^2-\frac{y^4}{32}\right) = 3y^2-\frac{15y^4}{32}. $$
Step 3 · Integrate over $y$
$$ V=\int_0^2\left(3y^2-\frac{15y^4}{32}\right)dy = \left[y^3-\frac{3y^5}{32}\right]_0^2 = 8-\frac{3\cdot32}{32} = 8-3 = 5. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: direct evaluation of $\displaystyle\int_0^2\left(3y^2-\frac{15y^4}{32}\right)dy$ gives exactly $5$.
Result
The volume is: