Cylinder, Cone and Sphere Problems: Difficult Problems with Solutions

Solution:
The points [tex]A[/tex], [tex]B[/tex], [tex]C[/tex] lie on the circle of the lower base, and the generatrix [tex]AA_1[/tex] is perpendicular to the plane of the base, so [tex]AB\perp AA_1[/tex] and [tex]AC\perp AA_1[/tex]. Hence [tex]\angle BAC[/tex] is the plane angle of the dihedral angle: [tex]\angle BAC=60^{\circ}[/tex].
Each section through two generatrices is a rectangle whose second side is the height [tex]h=AA_1=BB_1=CC_1[/tex]:
[tex]AB\times h=192, \ AC\times h=252, \ S_{BCC_1B_1}=BC\times h[/tex]
By the law of cosines in [tex]\triangle ABC[/tex]:
[tex]BC^2=AB^2+AC^2-2AB\times AC\cos 60^{\circ}=AB^2+AC^2-AB\times AC[/tex]
Multiplying by [tex]h^2[/tex], we get
[tex](BC\times h)^2=192^2+252^2-192\times 252=36864+63504-48384=51984[/tex]
[tex]BC\times h=\sqrt{51984}=228[/tex]
The area of the section [tex]BCC_1B_1[/tex] is [tex]228[/tex] square inches.

Solution:
Let [tex]O[/tex] be the center of the sphere and [tex]k[/tex] the section circle with center [tex]K[/tex] and radius [tex]\rho[/tex]. If a point is equidistant from all points of a circle, then by the theorem of Pythagoras its orthogonal projection on the plane of the circle is also equidistant from them, so this projection is the center [tex]K[/tex].
This holds for [tex]O[/tex] and for [tex]A[/tex], hence the line [tex]OA[/tex] is perpendicular to the plane of [tex]k[/tex] at [tex]K[/tex]. For every point [tex]M[/tex] of [tex]k[/tex] we have [tex]MK\perp OA[/tex], so [tex]\rho=MK[/tex] is the altitude from [tex]M[/tex] in [tex]\triangle OAM[/tex] with sides [tex]OM=20[/tex], [tex]AM=13[/tex], [tex]OA=11[/tex]. By Heron's formula with semiperimeter [tex]p=22[/tex]:
[tex]S_{OAM}=\sqrt{22\times 2\times 9\times 11}=66,\qquad \rho=\frac{2S_{OAM}}{OA}=\frac{2\times 66}{11}=12[/tex]
The length of the circle is [tex]2\pi\rho=24\pi[/tex] in. (Check: [tex]OK=\sqrt{20^2-12^2}=16[/tex], [tex]AK=16-11=5[/tex], [tex]AM=\sqrt{12^2+5^2}=13[/tex].)

Solution:
Let the cone have apex [tex]M[/tex], base center [tex]O[/tex], base radius [tex]r[/tex], generatrix [tex]l[/tex] and height [tex]h[/tex], and let [tex]\rho[/tex] be the radius of the ball. The axial section through a diameter [tex]AB[/tex] of the base is the isosceles triangle [tex]ABM[/tex] with [tex]AB=2r[/tex], [tex]MA=MB=l[/tex] and height [tex]MO=h[/tex].
By symmetry the center of the ball lies on [tex]MO[/tex], so this plane cuts the ball in a great circle that touches [tex]AB[/tex], [tex]MA[/tex] and [tex]MB[/tex], i.e. in the inscribed circle of [tex]\triangle ABM[/tex]. The area of [tex]\triangle ABM[/tex] is [tex]rh[/tex] and its semiperimeter is [tex]r+l[/tex], so [tex]\rho=\frac{rh}{r+l}[/tex].
The total surface area of the cone is [tex]S_1=\pi r^2+\pi rl=\pi r(r+l)[/tex], so for the volume [tex]V_1[/tex] of the cone:
[tex]V_1=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi r(r+l)\times \frac{rh}{r+l}=\frac{1}{3}S_1\rho[/tex]
For the ball, with surface area [tex]S_2=4\pi\rho^2[/tex] and volume [tex]V_2[/tex]:
[tex]V_2=\frac{4}{3}\pi\rho^3=\frac{1}{3}\times 4\pi\rho^2\times \rho=\frac{1}{3}S_2\rho\quad\Rightarrow\quad \frac{V_1}{V_2}=\frac{S_1}{S_2}=2[/tex]
Such a cone exists, e.g. [tex]r=\sqrt{2}\rho[/tex], [tex]h=4\rho[/tex]: then [tex]l=3\sqrt{2}\rho[/tex], [tex]\frac{rh}{r+l}=\rho[/tex] and [tex]S_1=8\pi\rho^2=2S_2[/tex]. So the volume of the cone is [tex]2[/tex] times the volume of the ball.

Solution:
Let [tex]R[/tex] be the radius of the ball. The volumes before and after the increase are [tex]\frac{4}{3}\pi R^3[/tex] and [tex]\frac{4}{3}\pi(R+6)^3[/tex], so
[tex]\frac{4}{3}\pi\left((R+6)^3-R^3\right)=1608\pi\Rightarrow (R+6)^3-R^3=\frac{3\times 1608}{4}=1206[/tex]
Expanding the cube:
[tex]18R^2+108R+216=1206\Rightarrow 18R^2+108R-990=0\Rightarrow R^2+6R-55=0[/tex]
The roots are [tex]R=5[/tex] and [tex]R=-11[/tex]. The radius is positive, so [tex]R=5[/tex] in. Check: [tex]\frac{4}{3}\pi\left(11^3-5^3\right)=\frac{4}{3}\pi\times 1206=1608\pi[/tex].

Solution:
The plane of each face cuts the sphere in a circle through the vertices of that face, so every face is inscribed in a circle. A lateral face is a parallelogram inscribed in a circle, hence a rectangle; so the lateral edges are perpendicular to the base and the prism is right.
The base [tex]ABCD[/tex] ([tex]AB\parallel CD[/tex]) is a cyclic trapezoid, hence isosceles, with angles [tex]30^{\circ}[/tex], [tex]30^{\circ}[/tex], [tex]150^{\circ}[/tex], [tex]150^{\circ}[/tex]. Its circumcircle (radius [tex]\rho[/tex]) is also the circumcircle of [tex]\triangle ABD[/tex], whose angle [tex]BAD[/tex] is [tex]30^{\circ}[/tex] or [tex]150^{\circ}[/tex]. By the law of sines
[tex]BD=2\rho\sin \angle BAD=2\rho\times \frac{1}{2}=\rho\Rightarrow \rho=4[/tex]
The center [tex]O[/tex] of the sphere is equidistant from the vertices of each base, so its projections on the planes of the bases are their circumcenters [tex]O_1[/tex] and [tex]O_2[/tex]. The prism is right, so [tex]O_1O_2[/tex] is perpendicular to both bases and [tex]O[/tex] lies on the line [tex]O_1O_2[/tex]. From the right triangles [tex]OO_1A[/tex] and [tex]OO_2A_1[/tex]
[tex]OO_1=OO_2=\sqrt{5^2-4^2}=3[/tex]
Since [tex]O_1\neq O_2[/tex], the point [tex]O[/tex] is the midpoint of [tex]O_1O_2[/tex], and the height of the prism is [tex]O_1O_2=3+3=6[/tex] in.

Solution:
Let [tex]P[/tex] be the common point of the chords, [tex]O[/tex] the center and [tex]R[/tex] the radius of the sphere. The plane of [tex]AB[/tex] and [tex]CD[/tex] cuts the sphere in a circle [tex]k[/tex] with center [tex]O_1[/tex] (the foot of the perpendicular from [tex]O[/tex]) and radius [tex]\rho[/tex]; from the right triangle [tex]OO_1A[/tex], [tex]R^2=\rho^2+OO_1^2=\rho^2+9[/tex].
By the intersecting chords theorem [tex]PA\times PB=PC\times PD[/tex], and [tex]PA+PB=PC+PD=4\sqrt{3}[/tex]; so [tex]PA[/tex], [tex]PB[/tex] and [tex]PC[/tex], [tex]PD[/tex] are the roots of the same quadratic equation and the two pairs coincide. Renaming [tex]C[/tex] and [tex]D[/tex] if necessary, [tex]PA=PC=x[/tex], [tex]PB=PD=y[/tex], [tex]x+y=4\sqrt{3}[/tex].
Let [tex]\theta=\angle APC=\angle BPD[/tex]; then [tex]\theta=60^{\circ}[/tex] or [tex]\theta=120^{\circ}[/tex]. If [tex]\theta=60^{\circ}[/tex], the isosceles triangles [tex]APC[/tex] and [tex]BPD[/tex] are equilateral, so [tex]AC=x\lt 8[/tex], [tex]BD=y\lt 8[/tex], and by the law of cosines [tex]AD^2=BC^2=x^2+y^2+xy=(x+y)^2-xy\lt 48[/tex]; no segment would be [tex]8[/tex]. Hence [tex]\theta=120^{\circ}[/tex], and by the law of cosines
[tex]AC=x\sqrt{3},\quad BD=y\sqrt{3},\quad AD^2=BC^2=x^2+y^2-xy\le\max(x,y)^2[/tex]
So the longest segment is the larger of [tex]AC[/tex] and [tex]BD[/tex]; let it be [tex]AC[/tex] ([tex]x\ge y[/tex]). Then
[tex]x\sqrt{3}=8\Rightarrow x=\frac{8\sqrt{3}}{3},\quad y=4\sqrt{3}-\frac{8\sqrt{3}}{3}=\frac{4\sqrt{3}}{3}[/tex]
[tex]BD=y\sqrt{3}=4,\quad AD^2=\frac{64}{3}+\frac{16}{3}-\frac{32}{3}=16\Rightarrow AD=4[/tex]
The circle [tex]k[/tex] is the circumcircle of [tex]\triangle ABD[/tex] with [tex]AD=BD=4[/tex], [tex]AB=4\sqrt{3}[/tex]. By the laws of cosines and sines
[tex]\cos\angle ADB=\frac{16+16-48}{2\times 4\times 4}=-\frac{1}{2}\Rightarrow \angle ADB=120^{\circ},\quad \rho=\frac{AB}{2\sin 120^{\circ}}=\frac{4\sqrt{3}}{\sqrt{3}}=4[/tex]
Then [tex]R^2=16+9=25[/tex], [tex]R=5[/tex], and the volume of the ball is
[tex]V=\frac{4}{3}\pi\times 5^3=\frac{500\pi}{3}[/tex]

Solution:
Let [tex]r[/tex] and [tex]h[/tex] be the radius and the height of the cylinder. From the lateral surface area
[tex]2\pi rh=80\pi\Rightarrow rh=40[/tex]
The plane is parallel to the axis, so the section is a rectangle [tex]ABB_1A_1[/tex], where [tex]AB[/tex] is a chord of the lower base and [tex]AA_1=BB_1=h[/tex] are generatrices. Let [tex]O[/tex] be the center of the lower base and [tex]K[/tex] the midpoint of [tex]AB[/tex]. Then [tex]OK\perp AB[/tex], and [tex]OK\perp AA_1[/tex] because the generatrices are perpendicular to the base, so [tex]OK[/tex] is perpendicular to the cutting plane. The axis is parallel to this plane, hence [tex]OK[/tex] is the distance from the axis to the plane: [tex]OK=4[/tex].
From the right triangle [tex]AKO[/tex] and from the diagonal of the rectangle
[tex]AK^2=r^2-16\Rightarrow AB^2=4(r^2-16), \quad AB^2+h^2=100[/tex]
Substituting [tex]h=\frac{40}{r}[/tex] we get
[tex]4r^2-64+\frac{1600}{r^2}=100\iff r^4-41r^2+400=0\iff r^2=16\text{ or }r^2=25[/tex]
If [tex]r=4[/tex], the plane is at distance [tex]r[/tex] from the axis, so it is tangent to the cylinder and the section is a single generatrix, not a rectangle. Hence [tex]r=5[/tex] and
[tex]h=\frac{40}{5}=8[/tex]
Check: [tex]AB=2\sqrt{25-16}=6[/tex] and [tex]\sqrt{6^2+8^2}=10[/tex].

Solution:
Let the cone have apex [tex]M[/tex], base center [tex]O[/tex] and base radius [tex]OA=r[/tex], and let the plane cut the height [tex]MO[/tex] at [tex]O_1[/tex] and the generatrix [tex]MA[/tex] at [tex]A_1[/tex]. The section is a circle with center [tex]O_1[/tex] and radius [tex]O_1A_1=r_1[/tex], and the plane cuts off a small cone with apex [tex]M[/tex] and this circle as its base.
Since [tex]O_1A_1\parallel OA[/tex], the right triangles [tex]MO_1A_1[/tex] and [tex]MOA[/tex] are similar, so the small cone is similar to the given one with ratio [tex]k=\frac{r_1}{r}=\frac{MO_1}{MO}[/tex]. The ratio of the areas of the section and the base is
[tex]\frac{\pi r_1^2}{\pi r^2}=k^2=\frac{1}{4}\Rightarrow k=\frac{1}{2}[/tex]
The volumes of similar solids are in the ratio [tex]k^3=\frac{1}{8}[/tex], so the small cone has volume [tex]264\times \frac{1}{8}=33[/tex]. The truncated cone is the rest of the given cone:
[tex]V=264-33=231[/tex]

Solution:
The semiperimeter of the triangle is [tex]p=33[/tex], and by Heron's formula its area is
[tex]S_{\triangle ABC}=\sqrt{33\times 3\times 8\times 22}=132[/tex]
Let [tex]CH[/tex] be the height to the line [tex]AB[/tex]. Then, from the right triangles [tex]AHC[/tex] and [tex]BHC[/tex],
[tex]CH=\frac{2\times 132}{11}=24, \quad AH=\sqrt{25^2-24^2}=7, \quad BH=\sqrt{30^2-24^2}=18[/tex]
Since [tex]BC^2=900\gt AB^2+AC^2=746[/tex], the angle [tex]BAC[/tex] is obtuse, so [tex]H[/tex] lies on the extension of [tex]BA[/tex] beyond [tex]A[/tex]; indeed [tex]BH=BA+AH=11+7=18[/tex].
During the rotation [tex]C[/tex] describes a circle with center [tex]H[/tex] and radius [tex]CH=24[/tex]. The segment [tex]BC[/tex] describes the lateral surface of a cone with apex [tex]B[/tex], height [tex]BH=18[/tex] and generatrix [tex]30[/tex], and [tex]AC[/tex] describes the lateral surface of a cone with apex [tex]A[/tex], height [tex]AH=7[/tex] and generatrix [tex]25[/tex], with the same base. The triangle [tex]ABC[/tex] is the triangle [tex]BHC[/tex] without the triangle [tex]AHC[/tex], so the solid is the big cone with the small cone removed:
[tex]V=\frac{1}{3}\pi\times 24^2\times 18-\frac{1}{3}\pi\times 24^2\times 7=\frac{1}{3}\pi\times 576\times 11=2112\pi[/tex]
The surface of the solid consists of the lateral surfaces of both cones:
[tex]S=\pi\times 24\times 30+\pi\times 24\times 25=720\pi+600\pi=1320\pi[/tex]

Solution:
Let the pyramid be [tex]ABCDM[/tex] with base edge [tex]a[/tex] and height [tex]MO=H[/tex], where [tex]O[/tex] is the center of the square [tex]ABCD[/tex], and let the cylinder have height [tex]x[/tex] ([tex]0\lt x\lt H[/tex]) and radius [tex]\rho[/tex].
The plane of the upper base of the cylinder is parallel to the base of the pyramid at distance [tex]x[/tex] from it. It meets [tex]MO[/tex] at a point [tex]O'[/tex] with [tex]MO'=H-x[/tex] and cuts the pyramid in a square [tex]A'B'C'D'[/tex], similar to [tex]ABCD[/tex] with ratio [tex]\frac{MO'}{MO}=\frac{H-x}{H}[/tex]; so its side is [tex]\frac{a(H-x)}{H}[/tex].
The upper circle of the cylinder lies in this plane, so it touches a lateral face exactly when it touches the line in which this face cuts the plane, i.e. a side of [tex]A'B'C'D'[/tex]. Hence the upper circle is inscribed in the square [tex]A'B'C'D'[/tex], and
[tex]\rho=\frac{a(H-x)}{2H}[/tex]
The volume of the cylinder is
[tex]V=\pi\rho^2x=\frac{\pi a^2}{4H^2}(H-x)^2x=\frac{\pi a^2}{8H^2}(H-x)(H-x)\times 2x[/tex]
The positive numbers [tex]H-x[/tex], [tex]H-x[/tex], [tex]2x[/tex] have the constant sum [tex]2H[/tex], so by the inequality between the arithmetic and geometric means
[tex](H-x)(H-x)\times 2x\le\left(\frac{2H}{3}\right)^3[/tex]
with equality exactly when [tex]H-x=2x[/tex], i.e. [tex]x=\frac{H}{3}[/tex]. So the volume of the cylinder is the largest when its height is [tex]\frac{H}{3}[/tex], and then the height of the pyramid is [tex]3[/tex] times the height of the cylinder.

Solution:
Let [tex]M[/tex] be the apex, [tex]O[/tex] the center of the base, [tex]r[/tex] the radius and [tex]l[/tex] the generatrix, and let [tex]MA\perp MB[/tex] be the two generatrices. The arcs [tex]AB[/tex] are in the ratio [tex]2:1[/tex], so they measure [tex]240^{\circ}[/tex] and [tex]120^{\circ}[/tex], and [tex]\angle AOB=120^{\circ}[/tex]. By the law of cosines in the isosceles triangle [tex]AOB[/tex]
[tex]AB^2=r^2+r^2-2r^2\cos 120^{\circ}=3r^2[/tex]
Triangle [tex]AMB[/tex] is right-angled and isosceles ([tex]MA=MB=l[/tex]), so [tex]AB^2=2l^2[/tex]. Since [tex]MO[/tex] is the height, triangle [tex]AOM[/tex] is right-angled at [tex]O[/tex] and [tex]l^2=r^2+36[/tex]. Therefore
[tex]3r^2=2(r^2+36)\Rightarrow r^2=72[/tex]
The volume of the cone is
[tex]V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi\times 72\times 6=144\pi[/tex]

Solution:
Let [tex]r[/tex] be the radius of the base, [tex]l[/tex] the generatrix and [tex]h[/tex] the height. From the lateral surface area
[tex]\pi rl=136\pi\Rightarrow rl=136[/tex]
The axial section is an isosceles triangle with legs [tex]l[/tex] and base [tex]2r[/tex], so
[tex]2l+2r=50\Rightarrow l+r=25[/tex]
Hence [tex]r[/tex] and [tex]l[/tex] are the roots of [tex]t^2-25t+136=0[/tex], i.e. [tex]8[/tex] and [tex]17[/tex]. The generatrix is the hypotenuse of the right triangle formed by the height, a radius and a generatrix, so [tex]l\gt r[/tex], i.e. [tex]r=8[/tex], [tex]l=17[/tex], and from this right triangle
[tex]h=\sqrt{l^2-r^2}=\sqrt{289-64}=15[/tex]
The volume of the cone is
[tex]V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi\times 64\times 15=320\pi[/tex]

Solution:
Let [tex]O_A[/tex], [tex]O_B[/tex], [tex]O_C[/tex] be the centers of the spheres. All three spheres lie on the same side of the plane: two touching spheres on opposite sides would touch each other at a point of the plane, and their points of tangency with the plane would coincide.
Take the spheres with centers [tex]O_A[/tex] and [tex]O_B[/tex], [tex]r_A\ge r_B[/tex]. The radii [tex]O_AA[/tex] and [tex]O_BB[/tex] are perpendicular to the plane, so [tex]ABO_BO_A[/tex] is a right trapezoid (or a rectangle) with [tex]O_AA=r_A[/tex], [tex]O_BB=r_B[/tex] and [tex]O_AO_B=r_A+r_B[/tex] (the spheres touch externally). Let [tex]E[/tex] be the foot of the perpendicular from [tex]O_B[/tex] to [tex]O_AA[/tex]; then [tex]O_BE=AB=c[/tex] and [tex]O_AE=r_A-r_B[/tex]. From the right triangle [tex]O_AEO_B[/tex]
[tex]c^2=(r_A+r_B)^2-(r_A-r_B)^2=4r_Ar_B\Rightarrow c=2\sqrt{r_Ar_B}[/tex]
In the same way
[tex]a=2\sqrt{r_Br_C}, \quad b=2\sqrt{r_Cr_A}[/tex]
Then
[tex]bc=4r_A\sqrt{r_Br_C}=2r_A\times a\Rightarrow r_A=\frac{bc}{2a}[/tex]
and similarly
[tex]r_B=\frac{ca}{2b}, \quad r_C=\frac{ab}{2c}[/tex]

Solution:
Let [tex]AB=a[/tex] and [tex]AD=h[/tex]; since [tex]AD\perp(ABC)[/tex], [tex]AD[/tex] is the height of the pyramid. Let [tex]M[/tex] be the midpoint of [tex]BC[/tex]. Then [tex]AM\perp BC[/tex] (a median of the equilateral triangle), and by the theorem of three perpendiculars [tex]DM\perp BC[/tex]. So [tex]\varphi=\angle AMD[/tex] is the angle between the planes [tex]BCD[/tex] and [tex]ABC[/tex].
Triangle [tex]ABC[/tex] is the orthogonal projection of triangle [tex]BCD[/tex] on the base, so
[tex]S_{ABC}=S_{BCD}\cos\varphi\Rightarrow \cos\varphi=\frac{\sqrt{2}}{2}\Rightarrow \varphi=45^{\circ}[/tex]
From the right triangle [tex]AMD[/tex] ([tex]\angle MAD=90^{\circ}[/tex])
[tex]h=AM\tan 45^{\circ}=AM=\frac{a\sqrt{3}}{2}[/tex]
For the faces: [tex]S_{ABC}=\frac{a^2\sqrt{3}}{4}[/tex], [tex]S_{BCD}=\sqrt{2}\,S_{ABC}[/tex], and the right triangles [tex]ABD[/tex] and [tex]ACD[/tex] with legs [tex]a[/tex] and [tex]h[/tex] have areas [tex]\frac{ah}{2}=\frac{a^2\sqrt{3}}{4}[/tex]. Hence the total surface area [tex]S[/tex] and the volume [tex]V[/tex] are
[tex]S=\frac{a^2\sqrt{3}}{4}(3+\sqrt{2}), \quad V=\frac{1}{3}\times \frac{a^2\sqrt{3}}{4}\times \frac{a\sqrt{3}}{2}=\frac{a^3}{8}[/tex]
Joining the center of the inscribed sphere with the vertices splits the pyramid into four pyramids whose bases are the faces and whose heights equal the radius [tex]r=1[/tex], so [tex]V=\frac{1}{3}Sr[/tex] and
[tex]1=\frac{3V}{S}=\frac{3a^3}{8}\times \frac{4}{a^2\sqrt{3}(3+\sqrt{2})}=\frac{a\sqrt{3}}{2(3+\sqrt{2})}\Rightarrow h=\frac{a\sqrt{3}}{2}=3+\sqrt{2}[/tex]

Solution:
Let the triangle be [tex]ABC[/tex] with [tex]AB=c[/tex], rotated about the line [tex]AB[/tex], and let [tex]CH=h[/tex] be its height to [tex]AB[/tex] ([tex]H[/tex] lies on the line [tex]AB[/tex]). Then
[tex]S=\frac{ch}{2}\Rightarrow h=\frac{2S}{c}[/tex]
During the rotation the point [tex]C[/tex] describes a circle with center [tex]H[/tex] and radius [tex]h[/tex], and the sides [tex]AC[/tex] and [tex]BC[/tex] describe the lateral surfaces of cones with apexes [tex]A[/tex] and [tex]B[/tex] whose base is the disc bounded by this circle.
If [tex]H[/tex] lies between [tex]A[/tex] and [tex]B[/tex], the solid consists of two cones with a common base of radius [tex]h[/tex] and heights [tex]AH[/tex] and [tex]HB[/tex]:
[tex]V=\frac{1}{3}\pi h^2\times AH+\frac{1}{3}\pi h^2\times HB=\frac{1}{3}\pi h^2(AH+HB)=\frac{1}{3}\pi h^2c[/tex]
If [tex]H[/tex] coincides with [tex]A[/tex] (right angle at the vertex [tex]A[/tex]), the solid is one cone with radius [tex]h[/tex] and height [tex]AB=c[/tex], so again [tex]V=\frac{1}{3}\pi h^2c[/tex].
If [tex]H[/tex] lies on the extension of [tex]AB[/tex] beyond [tex]A[/tex] (obtuse angle at the vertex [tex]A[/tex]; the case beyond [tex]B[/tex] is symmetric), the solid is the cone with apex [tex]B[/tex] and height [tex]HB[/tex] from which the cone with apex [tex]A[/tex] and height [tex]HA[/tex] is removed:
[tex]V=\frac{1}{3}\pi h^2\times HB-\frac{1}{3}\pi h^2\times HA=\frac{1}{3}\pi h^2(HB-HA)=\frac{1}{3}\pi h^2c[/tex]
In all cases, using [tex]h=\frac{2S}{c}[/tex], we get
[tex]V=\frac{1}{3}\pi h^2c=\frac{1}{3}\pi\times \frac{4S^2}{c^2}\times c=\frac{4\pi S^2}{3c}[/tex]

Solution:
Let the sphere have center [tex]O[/tex] and radius [tex]R[/tex], and let the truncated cone have base radii [tex]r_1\ne r_2[/tex] and generatrix [tex]l[/tex]. By symmetry [tex]O[/tex] lies on the axis of the cone. The axial section is an isosceles trapezoid [tex]ABCD[/tex] with bases [tex]AB=2r_1[/tex], [tex]CD=2r_2[/tex] and legs [tex]AD=BC=l[/tex], and the great circle of the sphere in this plane is inscribed in the trapezoid.
By symmetry this circle touches [tex]AB[/tex] at its midpoint [tex]O_1[/tex], [tex]CD[/tex] at its midpoint [tex]O_2[/tex], and [tex]BC[/tex] at a point [tex]T[/tex]. Tangent segments from a point to a circle are equal, so [tex]BT=BO_1=r_1[/tex] and [tex]CT=CO_2=r_2[/tex], hence
[tex]l=BC=BT+TC=r_1+r_2[/tex]
The center [tex]O[/tex] lies on the bisectors of the angles [tex]ABC[/tex] and [tex]BCD[/tex], whose sum is [tex]180^{\circ}[/tex] because [tex]AB\parallel CD[/tex]. Hence [tex]\angle OBC+\angle OCB=90^{\circ}[/tex], i.e. [tex]\angle BOC=90^{\circ}[/tex]. In the right triangle [tex]BOC[/tex] the segment [tex]OT[/tex] is the altitude to the hypotenuse, so
[tex]R^2=OT^2=BT\times TC=r_1r_2[/tex]
The area of the sphere and the lateral surface area of the truncated cone are
[tex]S_1=4\pi R^2=4\pi r_1r_2, \ S_2=\pi(r_1+r_2)l=\pi(r_1+r_2)^2[/tex]
[tex]S_2-S_1=\pi(r_1+r_2)^2-4\pi r_1r_2=\pi(r_1-r_2)^2\gt 0[/tex]
since [tex]r_1\ne r_2[/tex]. Hence the area of the sphere is less than the lateral surface area of the truncated cone.

Solution:
Let the bases of the truncated cone be the circles [tex]k_1[/tex] (center [tex]O_1[/tex], radius [tex]R[/tex]) and [tex]k_2[/tex] (center [tex]O_2[/tex], radius [tex]r[/tex]) in parallel planes; the axis [tex]O_1O_2[/tex] is perpendicular to both planes. A sphere is circumscribed about the truncated cone when both base circles lie on it.
Let a point [tex]Q[/tex] be equidistant from all points of [tex]k_1[/tex] and let [tex]Q'[/tex] be its orthogonal projection on the plane of [tex]k_1[/tex]. For every point [tex]X[/tex] of [tex]k_1[/tex] the triangle [tex]QQ'X[/tex] is right-angled, so [tex]Q'X^2=QX^2-QQ'^2[/tex] is the same for all [tex]X[/tex]. Hence [tex]Q'[/tex] is the center [tex]O_1[/tex] and [tex]Q[/tex] lies on the axis.
Conversely, if [tex]Q[/tex] lies on the axis, then for every point [tex]X[/tex] of [tex]k_1[/tex] the triangle [tex]QO_1X[/tex] is right-angled at [tex]O_1[/tex], so [tex]QX^2=QO_1^2+R^2[/tex] is the same for all [tex]X[/tex]; the same holds for [tex]k_2[/tex]. Therefore the center of every circumscribed sphere lies on the axis, and a point [tex]Q[/tex] of the axis is such a center if and only if it is equidistant from one point of [tex]k_1[/tex] and one point of [tex]k_2[/tex].
Consider the axial section, the isosceles trapezoid [tex]ABCD[/tex], where [tex]AB[/tex] is a diameter of [tex]k_1[/tex], [tex]CD[/tex] is a diameter of [tex]k_2[/tex] and [tex]AD[/tex] is a generatrix. A point [tex]Q[/tex] of the axis is the center of a circumscribed sphere if and only if [tex]QA=QD[/tex], i.e. [tex]Q[/tex] lies on the perpendicular bisector [tex]s[/tex] of [tex]AD[/tex] in the plane of the trapezoid (this plane contains the axis).
The points [tex]A[/tex] and [tex]D[/tex] lie in two different planes perpendicular to the axis, so [tex]AD[/tex] is not perpendicular to the axis. Hence [tex]s[/tex] is not parallel to the axis and meets it in exactly one point [tex]O[/tex].
Existence: by the above, the sphere with center [tex]O[/tex] and radius [tex]OA=OD[/tex] passes through all points of [tex]k_1[/tex] and [tex]k_2[/tex], so it is circumscribed about the truncated cone. Uniqueness: every circumscribed sphere has its center [tex]Q[/tex] on the axis with [tex]QA=QD[/tex], so [tex]Q=O[/tex] and its radius is [tex]OA[/tex]. (The point [tex]O[/tex] is the center of the circle circumscribed about the trapezoid [tex]ABCD[/tex].)