Informative line

### 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 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 = $$V(S_i)\approx A(x_i^*)\,\Delta x$$

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

as $$n\to\infty$$

$$V=\int\limits_a^b\,A(x)\,dx$$  #### 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 In this case the region when is rotated is given by (shaded region) Area of cross section at $$\therefore\,V=\int\limits_0^1\,\pi(x^2-x^6)\,dx$$

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

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

$$y^2=2y$$

$$\Rightarrow\,y=0,\,2$$ Area of cross section at

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

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

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

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

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

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

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

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

$$=\dfrac {64\,\pi}{15}$$ ### 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 Right Circular Cone without Integration

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

$$V=\dfrac {1}{3}\,\pi r^2\,h$$  #### 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

#### 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  Area of cross section $$=P_n=\pi×(2x)^2=4\,\pi x^2$$ $$\therefore$$ $$V=\int\limits_0^2\;4\,\pi x^2\;dx= 4\pi\dfrac {x^3}{3}\Bigg]_0^2=\dfrac {32\,\pi}{3}$$ ### 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 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 = $$V(S_i)\approx A(x_i^*)\,\Delta x$$

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

as $$n\to\infty$$

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

$$V=\int\limits_a^b\,A(x)\;dx$$ Area of cross section or

$$P_x=\pi×(\sqrt {x-2})^2$$

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

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

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

• 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 = $$V(S_i)\approx A(x_i^*)\,\Delta x$$

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

as $$n\to\infty$$

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

$$x=-1$$ 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 $$=\pi\,\Big( \text {( outer radius)}^2- \text {( inner radius)}^2\Big)$$

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

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

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

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

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

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

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

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

$$=\dfrac {29\,\pi}{30}$$ ### 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 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 $$v=\int \limits^a_0\,A(x)\;dx$$

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

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

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

$$=\dfrac{2\,a^3}{3}\pi$$

#### 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. Area of cross section is $$A(x)=\pi\,\left(1^2-x^2\right)$$ Volume $$V=\int \limits^1_0A(x)\,dx$$

When $$A(x)=\text{Area of cross section}$$ $$V=\int \limits^1_0\,\pi\left(1-x^2\right)\,dx$$

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

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

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

$$V=\dfrac{2\pi}{3}$$ ### 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

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

#### 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 $$x=-1$$. The cross section will be a washer with inner radius $$=y+1$$ and outer radius = 3. $$\therefore$$ 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