Volume Between a Paraboloid and a Plane, Outside a Cone: 04-BS-1, May 2014, 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. Paraboloid $z_p(r)=\tfrac74+\tfrac{r^2}{4}$ (with $r^2=x^2+y^2$); plane $z=4$; cone $z=2r$ (the upper nappe of $z^2=4r^2$).

Find. The volume of the paraboloid-to-plane region that lies outside the cone (i.e. where $z<2r$).

$(r,z)$ profile: paraboloid and cone cross at $r=1$ (both $z=2$); paraboloid meets the plane $z=4$ at $r=3$; the "outside the cone" region is bounded below by the paraboloid and above by $\min(2r,4)$, for $r\in[1,3]$.

Approach. First locate where the paraboloid crosses the cone and the plane. For $r$ below the first crossing, the paraboloid sits entirely above the cone, so no part of that slice is "outside" it. Beyond that crossing, integrate the annular ring whose height runs from the paraboloid up to whichever of {cone, plane} is lower, splitting at the point where the cone itself reaches the plane.

Locate the crossings. Paraboloid meets plane: $4=\tfrac74+\tfrac{r^2}{4}\Rightarrow r^2=9\Rightarrow r=3$. Paraboloid meets cone: $2r=\tfrac74+\tfrac{r^2}{4}\Rightarrow r^2-8r+7=0=(r-1)(r-7)\Rightarrow r=1$ (the relevant root, since $r=7$ is outside the $[0,3]$ range of interest). Cone meets plane: $2r=4\Rightarrow r=2$.
Identify the outside-cone region. For $r<1$: $z_p(1)=2$ at $r=1$; checking $r=0$ gives $z_p=1.75>0=$cone, so the paraboloid is above the cone for $r<1$ — that whole slice (from $z_p$ up to the plane) is entirely "inside" the cone ($z\ge2r$ throughout), contributing nothing. For $1
Set up the volume integral. $$V=2\pi\left[\int_1^2 r\big(2r-z_p(r)\big)\,dr+\int_2^3 r\big(4-z_p(r)\big)\,dr\right],\qquad z_p(r)=\tfrac74+\tfrac{r^2}{4}.$$
Evaluate the first integral. $$\int_1^2\Big(2r^2-\tfrac74r-\tfrac{r^3}{4}\Big)dr=\Big[\tfrac{2r^3}{3}-\tfrac{7r^2}{8}-\tfrac{r^4}{16}\Big]_1^2=\tfrac56-\Big(-\tfrac{13}{48}\Big)=\tfrac{53}{48}.$$
Evaluate the second integral. $$\int_2^3\Big(\tfrac94r-\tfrac{r^3}{4}\Big)dr=\Big[\tfrac{9r^2}{8}-\tfrac{r^4}{16}\Big]_2^3=\tfrac{81}{16}-\tfrac{56}{16}=\tfrac{25}{16}=\tfrac{75}{48}.$$ Summing: $\tfrac{53}{48}+\tfrac{75}{48}=\tfrac{128}{48}=\tfrac83$.

$$V=2\pi\cdot\frac83=\boxed{\dfrac{16\pi}{3}}\approx16.76$$

Quantity	Result
Paraboloid–cone crossing	$r=1$
Paraboloid–plane crossing	$r=3$
Cone–plane crossing	$r=2$
Volume outside the cone	$16\pi/3\approx16.76$

Concept & Theory

Topic. Volume of an axisymmetric solid between two surfaces, split by a third surface, via a radial cross-sectional profile.

Key relations. For an axisymmetric region described by $z$ between two functions of $r=\sqrt{x^2+y^2}$, the volume is $V=2\pi\int r\,[\,z_{\text{top}}(r)-z_{\text{bottom}}(r)\,]\,dr$. When a third surface splits the region, find every pairwise crossing radius first, then integrate piecewise between crossings using whichever surface is locally the binding upper/lower bound.

Why this works. Sketching the $(r,z)$ profile (rather than working in $3$D directly) turns this into an ordinary single-variable calculus problem: the three curves' pairwise intersections partition $r$ into intervals on which the ordering of surfaces (and hence the correct integrand) is fixed.

Common pitfall. Using the wrong upper bound for $r>2$ (forgetting that the cone has already exceeded the plane there, so the plane — not the cone — caps the slice), or missing the $r<1$ region entirely instead of showing it contributes zero.

Source. Stewart, Calculus: Early Transcendentals, Ch. 6 (§6.2 Volumes, disk/ring cross-sections) and 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)