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Volumes Of Solids Ii

Learn volume using integration of general solids, washer method and cylindrical shells method. Practice washer & shell method calculus formula.

Volume of a General Solid by Slicing Method

  • Let S be the solid that lies between \(x=a\) and \(x=b\). If the cross sectional area of \(S\) in plane \(P_x\) through \(x\) and \(\perp\) to x-axis is \(A(x)\) where \(A\) is a continuous function, then volume of \(S\)is given by \(V=\int\limits_a^b\,A(x)\;dx\)

  • We cut S into pieces and obtain a plane require that is called cross section of S. This cross sectional Area \(A(x)\) will vary as \(x\) increases function \('a'\) to \('b'\).
  • Divide S into \(x\) slabs of equal width \(\Delta x\) by using planes \(P_{x_1}, P_{x_2},....,P_{x_n}\) to slice the solid. The \(i^{th}\) slab is like a cylinder with base
  • Area = \(A(x_i^*)\,\big(x_i^*\in(x_{i-1}, \;x_i)\big)\) and height \(\Delta x\).

\(\therefore \) Volume of \(i^x\) slab = \(V(S_i)\approx A(x_i^*)\,\Delta x\)

Adding the volume 

\(V\approx\sum\limits_{i=1}^{n}\,A(x_i^*)\,\Delta x\)

 as \(n\to\infty\)

\(V=\int\limits_a^b\,A(x)\,dx\)

Illustration Questions

Find the volume of solid obtained by rotating the region bounded by the curves \(y=sin\,x,\;y=cos\,x\) and \(y=-1\) .\(\left ( 0\leq x \leq \dfrac {\pi}{4} \right)\)

A \(\pi\left ( \dfrac {4\sqrt 2-3}{2} \right)\)

B \(\pi\left ( \dfrac {\sqrt 2-1}{3} \right)\)

C \(\dfrac {22\pi}{3}\)

D \((\sqrt 2 + \sqrt 3)\pi\)

×

Let S be the solid that lies between \(x=a\) and \(x=b\). If the cross sectional area of \(S\) in plane \(P_x\) through \(x\) and \(\perp\) to x-axis is \(A(x)\) where, \(A\) is a continuous function, then volume of \(S\)is given by 

 \(V=\int\limits_a^b\,A(x)\;dx\)

In this case the region which is rotated is given by (shaded region). This region is to be rotated about the line 

\(y=-1\)

\(sin\,x=cos\,x\)

\(\Rightarrow x=\dfrac {\pi}{4}\)

image

The cross section will be a washer with inner radius

\(=1+sin\,x\) and outer radius \(=1+cos\,x\) 

\(\therefore\) Area of cross section \(=\pi\,\Big( \text {( outer radius)}^2- \text {( inner radius)}^2\Big)\)

\(=\pi\Big[(1+cos\,x)^2-(1+sin\,x)^2\Big]\)

\(=\pi\Big[1+cos^2\,x+2\,cos\,x-1-sin^2x-2\,sin\,x\Big]\)

\(=\pi\Big[cos^2\,x+2\,cos\,x-sin^2x-2\,sin\,x\Big]\)

\(=\pi\Big[cos\,2x+2\,cos\,x-2\,sin\,x\Big]\)

image

\(\therefore\,V=\int\limits_0^{\pi/4}\,\pi\Big( cos\,2x+2\,cos\,x-2\,sin\,x \Big)\,dx\)

\(=\pi\int\limits_0^{1}\,\Big( cos\,2x+2\,cos\,x-2\,sin\,x \Big)\,dx\)

\(=\pi \Big[ \dfrac {sin\,2x}{2}+2\,sinx+2\,cos \Big]_0^{\pi/4}\)

\(=\pi \left [\left( \dfrac {sin\,\pi/2}{2}+2\,sin\dfrac {\pi}{4}+2\,cos\dfrac {\pi}{4} \right)-(0+0+2) \right]\)

\(=\pi\left [ \dfrac {1}{2}+2\sqrt 2-2\right]\)

\(=\pi\left ( \dfrac {4\sqrt 2-3}{2}\right)\)

image

Find the volume of solid obtained by rotating the region bounded by the curves \(y=sin\,x,\;y=cos\,x\) and \(y=-1\) .\(\left ( 0\leq x \leq \dfrac {\pi}{4} \right)\)

A

\(\pi\left ( \dfrac {4\sqrt 2-3}{2} \right)\)

.

B

\(\pi\left ( \dfrac {\sqrt 2-1}{3} \right)\)

C

\(\dfrac {22\pi}{3}\)

D

\((\sqrt 2 + \sqrt 3)\pi\)

Option A is Correct

Volume of a General Solid by Slicing Method

  • Let S be the solid that lies between \(x=a\) and \(x=b\). If the cross sectional area of \(S\) in plane \(P_x\) through \(x\) and \(\perp\) to x-axis is \(A(x)\) where, \(A\) is a continuous function, then volume of \(S\)is given by \(V=\int\limits_a^b\,A(x)\;dx\)

  • We cut S into pieces and obtain a plane require that is called cross section of S. This cross sectional Area \(A(x)\) will vary as \(x\) increases function \('a'\) to \('b'\).
  • Divide S into \(x\) slabs of equal width \(\Delta x\) by using planes \(P_{x_1}, P_{x_2},....,P_{x_n}\) to slice the solid. The \(i^{th}\) slab is like a cylinder with base
  • Area = \(A(x_i^*)\,\big(x_i^*\in(x_{i-1}, \;x_i)\big)\) and height \(\Delta x\).

\(\therefore \) Volume of \(i^x\) slab = \(V(S_i)\approx A(x_i^*)\,\Delta x\)

Adding the volume 

\(V\approx\sum\limits_{i=1}^{n}\,A(x_i^*)\,\Delta x\)

 as \(n\to\infty\)

\(V=\int\limits_a^b\,A(x)\,dx\)

Illustration Questions

Consider the figure below and the various region marked on it. Find the volume of solid generated by rotating \(R_1\) about \(BC\).

A \(\dfrac {128\,\pi}{3}\)

B \(\dfrac {5\,\pi}{2}\)

C \(\dfrac {56\pi}{3}\)

D \(\dfrac {512\,\pi}{7}\)

×

Let S be the solid that lies between \(x=a\) and \(x=b\). If the cross sectional area of \(S\) in plane \(P_x\) through \(x\) and \(\perp\) to x-axis is \(A(x)\) where, \(A\) is a continuous function, then volume of \(S\)is given by 

 \(V=\int\limits_a^b\,A(x)\;dx\)

In this case, \(R_1\) is the area to be rotated about \(BC\).

Equation of BC is \(y=4\) and that of line OB is \(y=x\).

image

Area of cross section

\(=\pi\Big[(4)^2-(y)^2\Big]\)

\(=\pi\Big(16-y^2\Big)\)

image

\(\therefore\,V=\int\limits_0^{4}\,\pi\Big( 16-y^2\Big)\,dy\)

\(=\pi\int\limits_0^{4}\,\Big( 16-y^2 \Big)\,dy\)

\(=\pi \Big[ 16y-\dfrac {y^3}{3}\Big]_0^{4}\)

\(=\pi \Big( 64-\dfrac {64}{3}\Big)\)

\(=\pi ×\dfrac {2}{3}×64\)

\(=\dfrac {128\,\pi}{3}\)

image

Consider the figure below and the various region marked on it. Find the volume of solid generated by rotating \(R_1\) about \(BC\).

image
A

\(\dfrac {128\,\pi}{3}\)

.

B

\(\dfrac {5\,\pi}{2}\)

C

\(\dfrac {56\pi}{3}\)

D

\(\dfrac {512\,\pi}{7}\)

Option A is Correct

Volume by Cylindrical Shells

  • Some volumes are difficult to calculate by the method of disk or washers.
  • Consider the volume of solid obtained by rotating region bounded by \(y=0\)\(y=3x^2-x^3\) about y-axis.

  • Since \(x\)is difficult to evaluate in terms of \(y\) , the method of disks is not very useful.
  • Consider a cylindrical shell with inner radius \(r_1\) and outer radius \(r_2\).

Volume of shell = Volume of outer cylinder – Volume of inner cylinder

\(=V_2-V_1\\=\pi\,r_2^2\,h-\pi r_1^2\,h\)

\(=\pi\,h(r_2^2-r_1^2)\)

\(=\pi\,h(r_2+r_1)(r_2-r_1)\)

\(=2\pi\,h \left(\dfrac {r_1+r_2}{2}\right) (r_2-r_1)\)

\(=2\pi\,r\,h \;\Delta r\)

where,  \(\Delta r=r_2-r_1\)

\(\therefore\) Volume of cylindrical shell  \(=2\pi\,r\,h \;\Delta r\)

  • Now consider the solid obtained by rotating about y-axis, the region bounded by \(y=f(x) \Big(f(x) \geq 0 \Big)\) \(y=0\)\(x=a\) and \(x=b\) when \(b>a \geq 0\)

The volume of solid obtained is 

\(V=\displaystyle\int_a^b\,2\pi x\,f(x)\;dx\) where \(0\leq a <b\)

Cylindrical shell \(\rightarrow\;2\pi\,r\,h \;\Delta r\)

 

Illustration Questions

The volume of solid obtained by rotating the region bounded by \(y=x^3,\;y=0,\,x=2\) about y-axis by method of cylindrical shells is

A \(\dfrac {2\,\pi}{7}\)

B \(5\,\pi\)

C \(\dfrac {62\,\pi}{5}\)

D \(\dfrac {52\,\pi}{3}\)

×

The volume of solid obtained by rotating about y-axis, the region under the curve 

\(y=f(x)\) from \(a\) to \(b\) is

\(V=\displaystyle\int_a^b\,2\pi x\,f(x)\,dx\),  where \(0\leq a <b\)

In this case, the region to be rotated about y-axis is (shaded region)

image

\(\therefore\) Volume by cylindrical shells

\(V=\displaystyle\int_1^2\,2\pi x\,×x^3\;dx\)

\(=2\,\pi\displaystyle\int_1^2\,x^4\;dx\)

\(=2\,\pi\,\Big[\dfrac {x^5}{5}\Big]_1^2\)

\(=\dfrac {2\,\pi\,}{5}[32-1]\)

\(=\dfrac {62\,\pi\,}{5}\)

image

The volume of solid obtained by rotating the region bounded by \(y=x^3,\;y=0,\,x=2\) about y-axis by method of cylindrical shells is

A

\(\dfrac {2\,\pi}{7}\)

.

B

\(5\,\pi\)

C

\(\dfrac {62\,\pi}{5}\)

D

\(\dfrac {52\,\pi}{3}\)

Option C is Correct

Volume by Cylindrical Shells

  • Some volumes are difficult to calculate by the method of disk or washers.
  • Consider the volume of solid obtained by rotating region bounded by \(y=0\)\(y=3x^2-x^3\) about y-axis.

  • Since \(x\)is difficult to evaluate in terms of \(y\) , the method of disks is not very useful.
  • Consider a cylindrical shell with inner radius \(r_1\) and outer radius \(r_2\).

Volume of shell = Volume of outer cylinder – Volume of inner cylinder

\(=V_2-V_1=\pi\,r_2^2\,h-\pi r_1^2\,h\)

\(=\pi\,h(r_2^2-r_1^2)\)

\(=\pi\,h(r_2+r_1)(r_2-r_1)\)

\(=2\pi\,h \left(\dfrac {r_1+r_2}{2}\right) (r_2-r_1)\)

\(=2\pi\,r\,h \;\Delta r\)

where \(\Delta r=r_2-r_1\)

\(\therefore\) Volume of cylindrical shell  \(=\;\;2\pi\,r\,h \;\Delta r\)

  • Now consider the solid obtained by rotating about y-axis, the region bounded by \(y=f(x) \Big(f(x) \geq 0 \Big)\) \(y=0\)\(x=a\) and \(x=b\) when \(b>a \geq 0\)

The volume of solid obtained is 

\(V=\displaystyle\int_a^b\,2\pi x\,f(x)\;dx\) where \(0\leq a <b\)

Cylindrical shell \(\rightarrow \;\;2\pi\,r\,h \;\Delta r\)

 

Illustration Questions

Use method of cylindrical shells to find the volume generated by rotating the region bounded by \(y=x^2\) and \(y=6x-2x^2\) about y-axis.  

A \(5\,\pi\)

B \(8\,\pi\)

C \(\pi\)

D \(\dfrac {\pi}{24}\)

×

The volume of solid obtained by rotating about y-axis, the region under the curve 

\(y=f(x)\) from \(a\) to \(b\) is

\(V=\displaystyle\int_a^b\,2\pi x\,f(x)\;dx\) where \(0\leq a <b\)

image

In this case the region to be rotated about y-axis is (shaded region)

Intersection point 

\(x^2=6x-2x^2\)

\(3x^2-6x=0\)

\(\Rightarrow 3x(x-2)=0\)

\(x=0,\;x=2\)

image

\(\therefore\) Volume by cylindrical shells

\(V=\displaystyle\int_0^2\,2\pi x\,(6x-2x^2-x^2)\;dx\)

(Height of cylindrical shell) \(=6x-2x^2-x^2=6x-3x^2\)

\(\therefore \;V=2\,\pi\displaystyle\int_0^2\,(6x^2-3x^3)\;dx\)

\(=2\,\pi\, \left [ \dfrac {6x^3}{3}-\dfrac {3x^4}{4} \right]_0^2\)

\(=2\,\pi\, \left [ 2×8-\dfrac {3}{4}×16 \right]\)

\(=8\,\pi\,\)

image

Use method of cylindrical shells to find the volume generated by rotating the region bounded by \(y=x^2\) and \(y=6x-2x^2\) about y-axis.  

A

\(5\,\pi\)

.

B

\(8\,\pi\)

C

\(\pi\)

D

\(\dfrac {\pi}{24}\)

Option B is Correct

Volume by Cylindrical Shells

  • Some volumes are difficult to calculate by the method of disk or washers.
  • Consider the volume of solid obtained by rotating region bounded by \(y=0\)\(y=3x^2-x^3\) about y-axis.

  • Since \(x\)is difficult to evaluate in terms of \(y\) , the method of disks is not very useful.
  • Consider a cylindrical shell with inner radius \(r_1\) and outer radius \(r_2\).

Volume of shell = Volume of outer cylinder – Volume of inner cylinder

\(=V_2-V_1\\=\pi\,r_2^2\,h-\pi r_1^2\,h\)

\(=\pi\,h(r_2^2-r_1^2)\)

\(=\pi\,h(r_2+r_1)(r_2-r_1)\)

\(=2\pi\,h \left(\dfrac {r_1+r_2}{2}\right) (r_2-r_1)\)

\(=2\pi\,r\,h \;\Delta r\)

where,  \(\Delta r=r_2-r_1\)

\(\therefore\) Volume of cylindrical shell  \(=2\pi\,r\,h \;\Delta r\)

  • Now consider the solid obtained by rotating about y-axis, the region bounded by \(y=f(x) \Big(f(x) \geq 0 \Big)\) \(y=0\)\(x=a\) and \(x=b\) when \(b>a \geq 0\)

 

The volume of solid obtained is 

\(V=\displaystyle\int_a^b\,2\pi x\,f(x)\;dx\) where \(0\leq a <b\)

Cylindrical shell \(\rightarrow\;2\pi\,r\,h \;\Delta r\)

 

 

Illustration Questions

Use method of cylindrical shells to find the volume generated by rotating the region bounded by \(y=x^4,\;y=0,\,x=1\) about \(x=2\). 

A \(\dfrac {7\,\pi}{15}\)

B \(\dfrac {\pi}{2}\)

C \(\dfrac {5\,\pi}{3}\)

D \(\dfrac {6\,\pi}{5}\)

×

The volume of solid obtained by rotating about y-axis, the region under the curve 

\(y=f(x)\) from \(a\) to \(b\) is

\(V=\displaystyle\int_a^b\,2\pi x\,f(x)\,dx\),  where \(0\leq a <b\)

image

In this case, the region to be rotated about \(x=2\) is given by (shaded region)

image

Radius of cylindrical shell = \(2-x\)

Height of cylindrical shell = \(x^4\)

image

\(\therefore\) Volume by cylindrical shells

\(V=\displaystyle\int_0^1\,2\pi\,(2-x)\,x^4\;dx\)

\(=2\pi\displaystyle\int_0^1\,(2x^4-x^5)\;dx\)

\(=2\,\pi\,\Bigg[\dfrac {2x^5}{5}-\dfrac {x^6}{6}\Bigg]_0^1\)

\(=2\,\pi\,\Bigg[\dfrac {2}{5}-\dfrac {1}{6}\Bigg]\)

\(=2\,\pi\,\Bigg[\dfrac {12-5}{30}\Bigg]\)

\(=\dfrac {14\,\pi\,}{30}\)

\(=\dfrac {7\,\pi\,}{15}\)

image

Use method of cylindrical shells to find the volume generated by rotating the region bounded by \(y=x^4,\;y=0,\,x=1\) about \(x=2\). 

A

\(\dfrac {7\,\pi}{15}\)

.

B

\(\dfrac {\pi}{2}\)

C

\(\dfrac {5\,\pi}{3}\)

D

\(\dfrac {6\,\pi}{5}\)

Option A is Correct

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