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Volu= me and Solids of Revolution II
Following are more examples of solid whose cross–sectional area are tractable so that we may use t= he = Cavaleri’s principle to find their volumes.=
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Volu= me and Solids of Revolution II
Following are more examples of solid whose cross–sectional area are tractable so that we may use t= he = Cavaleri’s principle to find their volumes.=
One group of such solids consists the ones whose = cross–sectional areas are formed from the cross–sectio= ns of the base regions of the solids.
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Volu= me and Solids of Revolution II
Following are more examples of solid whose cross–sectional area are tractable so that we may use t= he = Cavaleri’s principle to find their volumes.=
One group of such solids consists the ones whose = cross–sectional areas are formed from the cross–sectio= ns of the base regions of the solids.
Example A.
A solid has a semi–circular = disc of radius r as base.
r
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Volu= me and Solids of Revolution II
Following are more examples of solid whose cross–sectional area are tractable so that we may use t= he = Cavaleri’s principle to find their volumes.=
One group of such solids consists the ones whose = cross–sectional areas are formed from the cross–sectio= ns of the base regions of the solids.
Example A.
A solid has a semi–circular = disc of radius r as base. Its cross sections, <= span style=3D'position:absolute;top:72.5%;left:5.24%;width:50.74%;height:6.0%'>= perpendicular to the = diameter of the base, are = isosceles right triangles as shown. Find its volume.
r
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Volu= me and Solids of Revolution II
Following are more examples of solid whose cross–sectional area are tractable so that we may use t= he = Cavaleri’s principle to find their volumes.=
One group of such solids consists the ones whose = cross–sectional areas are formed from the cross–sectio= ns of the base regions of the solids.
Example A.
A solid has a semi–circular = disc of radius r as base. Its cross sections, <= span style=3D'position:absolute;top:72.5%;left:5.24%;width:50.74%;height:6.0%'>= perpendicular to the = diameter of the base, are = isosceles right triangles as shown. Find its volume.
r
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Volu= me and Solids of Revolution II
Following are more examples of solid whose cross–sectional area are tractable so that we may use t= he = Cavaleri’s principle to find their volumes.=
One group of such solids consists the ones whose = cross–sectional areas are formed from the cross–sectio= ns of the base regions of the solids.
Example A.
A solid has a semi–circular = disc of radius r as base. Its cross sections, <= span style=3D'position:absolute;top:72.5%;left:5.24%;width:50.74%;height:6.0%'>= perpendicular to the = diameter of the base, are = isosceles right triangles as shown. Find its volume.
r
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Volu= me and Solids of Revolution II
Following are more examples of solid whose cross–sectional area are tractable so that we may use t= he = Cavaleri’s principle to find their volumes.=
One group of such solids consists the ones whose = cross–sectional areas are formed from the cross–sectio= ns of the base regions of the solids.
Example A.
A solid has a semi–circular = disc of radius r as base. Its cross sections, <= span style=3D'position:absolute;top:72.5%;left:5.24%;width:50.74%;height:6.0%'>= perpendicular to the = diameter of the base, are = isosceles right triangles as shown. Find its volume.
r
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Volu= me and Solids of Revolution II
We calculate the volume of half of the solid. Set the = x measurement as shown from x =3D 0= to x =3D r.
Given an x, the cross sectional len= gth is √r2 – x2.
x=3D0
x=3Dr
x
√= r2 – x2
So the cross section of the solid is a right isosceles = triangle with area ½ (√r2 – x2)2 =3D ½ = (r2 – x2).
So the volume of the entire solid (two halves) is
∫
x=3D0
r
dx
r2 – x2 
r
x=3D0
x=3Dr
x
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Volu= me and Solids of Revolution II
We calculate the volume of half of the solid. Set the = x measurement as shown from x =3D 0= to x =3D r.
Given an x, the cross sectional len= gth is √r2 – x2.
x=3D0
x=3Dr
x
√= r2 – x2
So the cross section of the solid is a right isosceles = triangle with area ½ (√r2 – x2)2 =3D ½ = (r2 – x2).
So the volume of the entire solid (two halves) is
∫
x=3D0
r
dx
r2 – x2 
=3D
r2x –=
x3 = ;
3
x=3D0
r
r
x=3D0
x=3Dr
x
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Volu= me and Solids of Revolution II
We calculate the volume of half of the solid. Set the = x measurement as shown from x =3D 0= to x =3D r.
Given an x, the cross sectional len= gth is √r2 – x2.
x=3D0
x=3Dr
x
√= r2 – x2
So the cross section of the solid is a right isosceles = triangle with area ½ (√r2 – x2)2 =3D ½ = (r2 – x2).
So the volume of the entire solid (two halves) is
∫
x=3D0
r
dx
r2 – x2 
=3D
r2x –=
x3 = ;
3
x=3D0
r
=3D
r
2r3 = ;
3
x=3D0
x=3Dr
x
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Δx
r
h
2πr
Δx
h
= ≈
Volu= me and Solids of Revolution II
Therefore the volume of the cylindrical shell is roughly
2πrhΔx.
x=3D= a
x=3D= b
y= =3Df(x)
Suppose a solid of revolution is formed by a template that is bounded by y =3D f(x), from x =3D a to
x =3D b, around a vertical axis.
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Δx
r
h
2πr
Δx
h
= ≈
Volu= me and Solids of Revolution II
Therefore the volume of the cylindrical shell is roughly
2πrhΔx.
x=3D= a
x=3D= b
y= =3Df(x)
Suppose a solid of revolution is formed by a template that is bounded by y =3D f(x), from x =3D a to
x =3D b, around a vertical axis.
x= =3Da
x=3D= b
y= =3Df(x)
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x=3D= a
x=3D= b
Partition [a, b] into n equal size subintervals&#= 13;
{x0=3Da= , x1= , x2, ..= xi= , .. xn=3Db= }, xi=  be = an arbitrary point in [xi= –1, xi]. L= et Δx be the width of the subinterval.
Then the shell generated by the strip over [xi= –1, xi] &#= 13;
is approximately the cylindrical shell with heigh= t f(xi= ), width Δx, and radius D(xi).
*
xi
*
*
Volu= me and Solids of Revolution II
*
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x=3D= a
x=3D= b
y= =3Df(xi)
Partition [a, b] into n equal size subintervals&#= 13;
{x0=3Da= , x1= , x2, ..= xi= , .. xn=3Db= }, xi=  be = an arbitrary point in [xi= –1, xi]. L= et Δx be the width of the subinterval.
Then the shell generated by the strip over [xi= –1, xi] &#= 13;
is approximately the cylindrical shell with heigh= t f(xi= ), width Δx, and radius D(xi).
*
xi
D(x= )
*
*
Volu= me and Solids of Revolution II
*
*
*
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x=3D= a
x=3D= b
y= =3Df(xi)
Partition [a, b] into n equal size subintervals&#= 13;
{x0=3Da= , x1= , x2, ..= xi= , .. xn=3Db= }, xi=  be = an arbitrary point in [xi= –1, xi]. L= et Δx be the width of the subinterval.
Then the shell generated by the strip over [xi= –1, xi] &#= 13;
is approximately the cylindrical shell with heigh= t f(xi= ), width Δx, and radius D(xi).
*
xi
D(x= )
*
*
x= =3Da
x=3D= b
D(x= i )
Volu= me and Solids of Revolution II
*
*
 *
*
y= =3Df(xi)
*
xi
*
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x=3D= a
x=3D= b
y= =3Df(xi)
Partition [a, b] into n equal size subintervals&#= 13;
{x0=3Da= , x1= , x2, ..= xi= , .. xn=3Db= }, xi=  be = an arbitrary point in [xi= –1, xi]. L= et Δx be the width of the subinterval.
Then the shell generated by the strip over [xi= –1, xi] &#= 13;
is approximately the cylindrical shell with heigh= t f(xi= ), width Δx, and radius D(xi).
*
xi
D(x= )
2πD(xi= )f(xi)Δx. <= /div>
*
*
x= =3Da
x=3D= b
xi
D(x= i )
*
*
Hence = the volume of this cylindrical shell is
Volu= me and Solids of Revolution II
*
*
 *
*
y= =3Df(xi)
*
*
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= 776; 2= πD(xi)f(x= i= ) = 16;x. <= /div>
*
*
Volu= me and Solids of Revolution II
x=3D= a
x=3D= b
y=3D= f(xi)
xi
D(x= )
x= =3Da
x=3D= b
D(x= i )
*
*
 *
*
y= =3Df(xi)
*=
xi
*
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= 776; 2= πD(xi)f(x= i= ) = 16;x. <= /div>
*
*
The volume of the solid of revolution V is the sum of the volumes of all such cylindrical shells, i.e. 
Volu= me and Solids of Revolution II
x=3D= a
x=3D= b
y=3D= f(xi)
xi
D(x= )
x= =3Da
x=3D= b
D(x= i )
*
*
 *
*
y= =3Df(xi)
*=
xi
*
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= 776; 2= πD(xi)f(x= i= ) = 16;x. <= /div>
*
*
Volume= of V
= 776;
i=3D= 1
n
ͨ= 1;
The volume of the solid of revolution V is the sum of the volumes of all such cylindrical shells, i.e. 
2πD(xi= )f(xi)Δx <= /div>
*
*
Volu= me and Solids of Revolution II
x=3D= a
x=3D= b
y=3D= f(xi)
xi
D(x= )
x= =3Da
x=3D= b
D(x= i )
*
*
 *
*
y= =3Df(xi)
*=
xi
*
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= 776; 2= πD(xi)f(x= i= ) = 16;x. <= /div>
*
*
Volume= of V
= 776;
i=3D= 1
n
ͨ= 1;
In fac= t the volume is its limit as = 16;xà0= or n &agrav= e;∞.
The volume of the solid of revolution V is the sum of the volumes of all such cylindrical shells, i.e. 
2πD(xi= )f(xi)Δx <= /div>
*
*
Volu= me and Solids of Revolution II
x=3D= a
x=3D= b
y=3D= f(xi)
xi
D(x= )
x= =3Da
x=3D= b
D(x= i )
*
*
 *
*
y= =3Df(xi)
*=
xi
*
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= 776; 2= πD(xi)f(x= i= ) = 16;x. <= /div>
*
*
Volume= of V
= 776;
i=3D= 1
n
ͨ= 1;
Theref= ore the volume of V =3D lim
i=3D= 1
n
W= 21;
n à∞
In fac= t the volume is its limit as = 16;xà0= or n &agrav= e;∞.
The volume of the solid of revolution V is the sum of the volumes of all such cylindrical shells, i.e. 
2πD(xi= )f(xi)Δx <= /div>
*
*
2πD(xi= )f(xi)Δx <= /div>
*
*
Volu= me and Solids of Revolution II
x=3D= a
x=3D= b
y=3D= f(xi)
xi
D(x= )
x= =3Da
x=3D= b
D(x= i )
*
*
 *
*
y= =3Df(xi)
*=
xi
*
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= 776; 2