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Volumes Of Solids I

Learn volume of a right circular cone & volumes of general solids formula for Integration. Practice washer 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 region that is called cross-section of S. This cross-sectional Area \(A(x)\) will vary as \(x\) increases from \('a'\) to \('b'\).
  • Divide S into \(n\) 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 the 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 \(y=\sqrt {x-2},\;x=4, \,y=0\) about x-axis.

A \(16\pi\)

B \(2\pi\)

C \(512\pi\)

D \(8\pi\)

×

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 

 \(V=\int\limits_a^b\,A(x)\;dx\)

image

Area of cross section or 

\(P_x=\pi×(\sqrt {x-2})^2\)

\(=\pi\,(x-2)\)

image

\(\therefore\,V=\int\limits_2^4\,\pi(x-2)\,dx\)

\(=\pi\Bigg[\left ( \dfrac {x^2}{2}-2x\right) \Bigg]_2^4\)

\(=\pi[(8-8)-(2-4)]\\=2\pi\)

image

Find the volume of solid obtained by rotating the region bounded by \(y=\sqrt {x-2},\;x=4, \,y=0\) about x-axis.

A

\(16\pi\)

.

B

\(2\pi\)

C

\(512\pi\)

D

\(8\pi\)

Option B 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 sectional 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  and  about x-axis.

A \(\dfrac{5\pi}{8}\)

B \(\dfrac{4\pi}{21}\)

C \(6\pi\)

D \(\dfrac{121\pi}{3}\)

×

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

 

image

In this case the region when is rotated is given by (shaded region)

 

 

image

Area of cross section at

 

 

image

\(\therefore\,V=\int\limits_0^1\,\pi(x^2-x^6)\,dx\)

\(=\pi\Bigg[ \dfrac {x^3}{3}- \dfrac {x^7}{7}\Bigg]_0^1\)

\(=\pi\Bigg[\dfrac {1}{3}- \dfrac {1}{7}\Bigg]\\=\dfrac{4\pi}{21}\)

image

Find the volume of solid obtained by rotating the region bounded by the curves  and  about x-axis.

A

\(\dfrac{5\pi}{8}\)

.

B

\(\dfrac{4\pi}{21}\)

C

\(6\pi\)

D

\(\dfrac{121\pi}{3}\)

Option B is Correct

Illustration Questions

Find the volume of solid obtained by rotating the region bounded by the curves \(y^2=x\) and \(x=2y\,\) about y-axis.

A \(\dfrac {64\pi}{15}\)

B \(\dfrac {\pi}{8}\)

C \(\dfrac {32\pi}{5}\)

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

×

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 the region which is rotated is given by (shaded region)

\(y^2=2y\)

\(\Rightarrow\,y=0,\,2\)

image

Area of cross section at

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

\(=\pi(4y^2-y^4)\)

 

image

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

\(=\pi\,\int\limits_0^2\,(4y^2-y^4)\,dy\)

\(=\pi\left [ \dfrac {4y^3}{3}-\dfrac {y^5}{5}\right] _0^2\)

\(=\pi\left [ \dfrac {4×8}{3}-\dfrac {32}{5}\right]\)

\(=\pi\left [ \dfrac {32}{3}-\dfrac {32}{5}\right]\)

\(=32\,\pi\left [ \dfrac {1}{3}-\dfrac {1}{5}\right]\)

\(=32\,\pi\left [ \dfrac {2}{15}\right]\)

\(=\dfrac {64\,\pi}{15}\)

image

Find the volume of solid obtained by rotating the region bounded by the curves \(y^2=x\) and \(x=2y\,\) about y-axis.

A

\(\dfrac {64\pi}{15}\)

.

B

\(\dfrac {\pi}{8}\)

C

\(\dfrac {32\pi}{5}\)

D

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

Option A is Correct

Volume of a General Solid by Slicing Method

  • Obtained by rotating a region through a line other than axis.
  • 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 sectional 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 \(y=x^2,\;x=y^2\) and \(x=-1\) line.

A \(\dfrac {\pi}{28}\)

B \(\dfrac {29\pi}{30}\)

C \(\dfrac {15}{16}\pi\)

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

×

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. 

\(x=-1\)

image

The cross section will be a washer with inner radius= \(1+y^2\) and

the outer radius \(=1+\sqrt y\)

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

\(=\pi\Big[(1+\sqrt y)^2-(1+y^2)^2\Big]\)

\(=\pi\Big[1+y+2\sqrt y-1-y^4-2y^2\Big]\)

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

image

\(\therefore\,V=\int\limits_0^1\,\pi\Big(y+2\sqrt y-y^4-2y^2\Big)\,dy\)

\(=\int\limits_0^1\,\pi\Big(y+2\sqrt y-y^4-2y^2\Big)\,dy\)

\(=\pi\left [ \dfrac {y^2}{2}+\dfrac{2y^{3/2}}{3/2}-\dfrac {y^5}{5}-\dfrac {2y^3}{3}\right] _0^1\)

\(=\pi\left [ \dfrac {1}{2}+\dfrac{4}{3}-\dfrac {1}{5}-\dfrac {2}{3}\right]\)

\(=\pi\left [ \dfrac {15+40-6-20}{30}\right]\)

\(=\pi×\left [ \dfrac {29}{30}\right]\)

\(=\dfrac {29\,\pi}{30}\)

image

Find the volume of solid obtained by rotating the region bounded by the curves \(y=x^2,\;x=y^2\) and \(x=-1\) line.

A

\(\dfrac {\pi}{28}\)

.

B

\(\dfrac {29\pi}{30}\)

C

\(\dfrac {15}{16}\pi\)

D

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

Option B is Correct

Volume of Right Circular Cone without Integration

  • The volume of a right circular cone whose radius is \(r\) and height is \(h\), is given by 

      \(V=\dfrac {1}{3}\,\pi r^2\,h\)

Illustration Questions

Find the volume of a cone whose radius is \(2m\) and height is \(5m\).

A \(\dfrac {10\,\pi}{3}\,m^3\)

B \(\dfrac {20\,\pi}{3}\,m^3\)

C \(5\,\pi\,m^3\)

D \(\dfrac {\pi}{3}\,m^3\)

×

Volume of a cone = \(\dfrac {1}{3}\,\pi r^2\,h=V\)

where 

\(r=\) radius of cone

\(h=\) height of cone

In this problem,

\(r=2\,m,\;h=5m\)

\(\therefore\) \(V=\dfrac {1}{3}\,\pi\,(2)^2\,×5\)

\(=\dfrac {20\,\pi}{3}\,m^3\)

Find the volume of a cone whose radius is \(2m\) and height is \(5m\).

A

\(\dfrac {10\,\pi}{3}\,m^3\)

.

B

\(\dfrac {20\,\pi}{3}\,m^3\)

C

\(5\,\pi\,m^3\)

D

\(\dfrac {\pi}{3}\,m^3\)

Option B is Correct

Volume of Right Circular Cone by Disk and Washer Method

  • The volume of a right circular cone whose radius is \(r\) and height is \(h\), is given by 

      \(V=\dfrac {1}{3}\,\pi r^2\,h\)

  • Consider the line  \(y=x×m\)  and the region bounded by \(y=x\) and \(x=a\) rotated about x-axis. The figure formed will be a right circular cone.

 

 

\(V=\displaystyle\int\limits_0^a\,\pi(mx)^2\;dx=\pi\,m^2\int\limits_0^a\,x^2\;dx\)

\(=\pi\,m^2\;\dfrac {x^3}{3}\Bigg]_0^a\,\)

\(=\pi\,m^2\;\dfrac {a^3}{3}\)

\(=\dfrac {\pi}{3}×a×(m\,a)^2\)

\(=\dfrac {1}{3}\,\pi\,r^2\,h\)

where 

\(r=\) radius of cone formed

\(h=\) height of cone

Illustration Questions

Find the volume of solid obtained by rotating region bounded by the line \(y=2x\) and \(x=2\,(x\geq0)\) about x-axis.

A \(\dfrac {16\,\pi}{3}\)

B \(\dfrac {32\,\pi}{3}\)

C \(\dfrac {4\,\pi}{3}\)

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

×

In this case the region which is rotated is given by shaded region

image image

Area of cross section \(=P_n=\pi×(2x)^2=4\,\pi x^2\)

 

image

\(\therefore \) \(V=\int\limits_0^2\;4\,\pi x^2\;dx= 4\pi\dfrac {x^3}{3}\Bigg]_0^2=\dfrac {32\,\pi}{3}\)

image

Find the volume of solid obtained by rotating region bounded by the line \(y=2x\) and \(x=2\,(x\geq0)\) about x-axis.

A

\(\dfrac {16\,\pi}{3}\)

.

B

\(\dfrac {32\,\pi}{3}\)

C

\(\dfrac {4\,\pi}{3}\)

D

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

Option B is Correct

Volume of General Solids by Shell Method 

Consider a region \(R\) enclosed by lines. \(x=0,\;y=x\) and \(y=a\) is rotated about \(x-\) axis.

Lines \(x=0,\;y=x\) and \(y=a\) intersect at the points \((0,\,0),\;(0,\,a)\;\&\;(a,\,a)\) as shown in figure.

Area of cross section is

\(A(R)=\pi\,\left(a^2-x^2\right)\) & volume \(v=\int \limits^a_0\,A(x)\;dx\)

\(V=\int \limits^a_0\,\pi\,\left(a^2-x^2\right)\,dx\)

\(=\pi\,\left[a^2[x]^a_0-\left[\dfrac{x^3}{3}\right]^a_0\right]\)

\(=\pi\,\left(a^3-\dfrac{a^3}{3}\right)\)

\(=\dfrac{2\,a^3}{3}\pi\)

Illustration Questions

The region \(R\) enclosed by lines \(x=0,\;y=x\) and \(y=1\) is rotated about the \(x-\) axis. Find the volume of the resulting solid.

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

B \(\pi\)

C \(\dfrac{2\pi}{3}\)

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

×

lines \(x=0,\;y=x\) and \(y=1\) intersect at points \((0,\,0),\;(0,\,1)\) and \((1,\,1)\) as shown in figure.

image

Area of cross section is \(A(x)=\pi\,\left(1^2-x^2\right)\)

image

Volume \(V=\int \limits^1_0A(x)\,dx\)

When \(A(x)=\text{Area of cross section}\)

image

\(V=\int \limits^1_0\,\pi\left(1-x^2\right)\,dx\)

\(=\pi\left[\,\int \limits^1_0\,dx-\int\limits^1_0\,x^2\,dx\right]\)

\(=\pi\,\left[[x]^1_0-\left[\dfrac{x^3}{3}\right]\right]^1_0\)

\(=\pi\,\left[1-\dfrac{1}{3}\right]\)

\(V=\dfrac{2\pi}{3}\)

image

The region \(R\) enclosed by lines \(x=0,\;y=x\) and \(y=1\) is rotated about the \(x-\) axis. Find the volume of the resulting solid.

A

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

.

B

\(\pi\)

C

\(\dfrac{2\pi}{3}\)

D

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

Option C is Correct

Volume of General Solids by Shell Method

When solids are obtained by revolving a region about a line, such solids are called as solids of revolution.

  • If the cross-section is a washer, inner radius \(r_{in}\)  and outer radius \(r_{out}\)  can be found from a sketch as shown in the figure.
  • Area of the washer can be found by subtracting the area of the inner disk from the area of outer disk.

\(A=\pi\,(\text{Outer radius})^2-\pi\,(\text{Inner radius})^2\)

Further volume can be calculated as

\(V=\int\limits^b_a\,A(y)\,dy\;or\;\int\limits^b_a\,A(x)\,dx\)

Illustration Questions

Find the volume of solid obtained by rotating the region bounded by \(x\) axis, \(y=x,\;x=2\) about \(x=-1\) line.

A \(\dfrac{16\pi}{3}\)

B \(\dfrac{28\pi}{3}\)

C \(\dfrac{16\pi}{5}\)

D \(\dfrac{11\pi}{3}\)

×

In this case the region which is rotated is shaded region. This region is to be rotated about \(x=-1\).

image

The cross section will be a washer with inner radius \(=y+1\) and outer radius = 3.

image

\(\therefore\) Area of cross section \(A(y)=\pi\,\left[(\text{outer radius})^2-(\text{inner radius})^2\right]\)

\(A(y)=\pi\,\left[3^2-(y+1)^2\right]\)

\(A(y)=\pi\,\left[9-y^2-1-2y\right]\)

\(A(y)=\pi\,\left[8-y^2-2y\right]\)

\(V=\int\limits^2_0\,A(y)\,dy\)

\(\displaystyle\therefore\;\text{Volume = V}=\int\limits^2_0\,\pi\,\left[8-y^2-2y\right]\,dy\)

\(\displaystyle=\pi\int\limits^2_0\,\left[8dy-\int\limits^2_0\,y^2dy-2\int\limits^2_0\,y\,dy\right]\)

\(=\pi\Bigg[8[y]^2_0-\left[\dfrac{y^3}{3}\right]^2_0-2\left[\dfrac{y^2}{2}\right]^2_0\)

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

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

\(=\pi\left[\dfrac{48-8-12}{3}\right]\)

\(=\pi\left[\dfrac{32}{3}\right]\)

Find the volume of solid obtained by rotating the region bounded by \(x\) axis, \(y=x,\;x=2\) about \(x=-1\) line.

A

\(\dfrac{16\pi}{3}\)

.

B

\(\dfrac{28\pi}{3}\)

C

\(\dfrac{16\pi}{5}\)

D

\(\dfrac{11\pi}{3}\)

Option B is Correct

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