Volume Between a Paraboloid and a Plane, Outside a Cone: 04-BS-1, May 2016, Q6

Question 6: Volume Between a Paraboloid and a Plane, Outside a Cone 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. In cylindrical coordinates ($r^2=x^2+y^2$): paraboloid $z=\tfrac74+\tfrac{r^2}{4}$, plane $z=4$, cone $z=2r$ (upper nappe).

Find. The volume of the part of the paraboloid–plane solid lying outside the cone (i.e. where $z<2r$).

Radial cross-section: the shaded solid outside the cone runs from $r=1$ (where cone meets paraboloid) to $r=3$ (where paraboloid meets the plane), bounded above by the cone (for $1\le r\le2$) or the plane (for $2\le r\le3$), below always by the paraboloid.

Approach. Find where the three surfaces cross in $r$, then integrate the annular-ring volume $2\pi r\,[\text{top}(r)-\text{bottom}(r)]\,dr$ over each radial band, using whichever of the cone or the plane forms the "outside-the-cone" upper bound in that band.

Paraboloid–plane intersection. $4=\tfrac74+\tfrac{r^2}{4}\Rightarrow r^2=9\Rightarrow r=3$. So the bounded solid (paraboloid below, plane above) exists for $0\le r\le3$.
Paraboloid–cone intersection. $2r=\tfrac74+\tfrac{r^2}{4}\Rightarrow r^2-8r+7=0\Rightarrow(r-1)(r-7)=0\Rightarrow r=1$ (the root $r=7$ lies outside the relevant range $[0,3]$).
Identify "outside the cone" by band. For $0\le r\le1$: the paraboloid stays above the cone, since $\tfrac74+\tfrac{r^2}{4}-2r=\tfrac14(r-1)(r-7)>0$ for $r<1$ (e.g. at $r=\tfrac12$ the paraboloid is $\tfrac{29}{16}=1.8125$ against the cone's $1$), so the entire solid at that $r$ sits above the cone (inside it) — no contribution. For $1\le r\le2$: the cone $2r$ rises from $2$ to $4$, staying between the paraboloid and the plane, so the outside-cone slice runs from the paraboloid up to the cone. For $2\le r\le3$: the cone height $2r\ge4$ exceeds the plane, so the entire slice from paraboloid to plane is outside the cone.
Integrate the two bands. $$V=\int_1^2 2\pi r\left[2r-\left(\tfrac74+\tfrac{r^2}{4}\right)\right]dr+\int_2^3 2\pi r\left[4-\left(\tfrac74+\tfrac{r^2}{4}\right)\right]dr.$$ Evaluating (each is a routine polynomial integral in $r$): $$\int_1^2 2\pi r\cdot\tfrac14(r-1)(7-r)\,dr=\frac{53\pi}{24},\qquad \int_2^3 2\pi r\left(\tfrac94-\tfrac{r^2}{4}\right)dr=\frac{25\pi}{8}.$$ (The first bracket factors as $2r-\tfrac74-\tfrac{r^2}{4}=-\tfrac14\!\left(r^2-8r+7\right)=\tfrac14(r-1)(7-r)$, which vanishes at $r=1$ as it must.)
Sum the two bands. $$V=\frac{53\pi}{24}+\frac{25\pi}{8}=\frac{53\pi}{24}+\frac{75\pi}{24}=\frac{128\pi}{24}=\boxed{\dfrac{16\pi}{3}}\approx16.76$$

Quantity	Result
Paraboloid–cone crossing	$r=1$
Paraboloid–plane crossing	$r=3$
Band $1\le r\le2$ volume	$53\pi/24$
Band $2\le r\le3$ volume	$25\pi/8$
Total volume $V$	$16\pi/3\approx16.76$

Concept & Theory

Topic. Volume of a solid of revolution bounded by multiple axisymmetric surfaces, via annular-ring (washer) integration.

Key relations. For axisymmetric surfaces $z=h(r)$, the washer volume element is $dV=2\pi r\,[\text{top}(r)-\text{bottom}(r)]\,dr$, since each radius $r$ sweeps a thin annular ring of circumference $2\pi r$ through the full $2\pi$ of $\theta$.

Why this works. All three surfaces (paraboloid, plane, cone) depend only on $r=\sqrt{x^2+y^2}$, so the "outside the cone" condition $z<2r$ can be tracked purely as a function of $r$ — the upper bound of integration switches from the cone to the plane exactly where the cone surface itself crosses the plane, at $r=2$ (since $2r=4$ there).

Common pitfall. Integrating the outside-cone volume over the wrong radial range (e.g. using $[0,3]$ throughout instead of recognizing that $[0,1]$ contributes nothing); forgetting the factor of $r$ in the volume element; missing that the upper bound switches at $r=2$, not at the paraboloid–cone crossing $r=1$.

Source. Stewart, Calculus: Early Transcendentals, Ch. 15 (§15.7 Triple Integrals in Cylindrical Coordinates).

Question 6 of 8: Volume Between a Paraboloid and a Plane, Outside a Cone

Question 6: Volume Between a Paraboloid and a Plane, Outside a Cone 20 marks