Informative line

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

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

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 $$y=sin\,x,\;y=cos\,x$$ 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

$$y=-1$$

$$sin\,x=cos\,x$$

$$\Rightarrow x=\dfrac {\pi}{4}$$

The cross section will be a washer with inner radius

$$=1+sin\,x$$ and outer radius $$=1+cos\,x$$

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

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

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

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

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

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

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

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

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

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

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

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

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

as $$n\to\infty$$

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

#### 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 $$y=x$$.

Area of cross section

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

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

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

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

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

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

$$=\pi ×\dfrac {2}{3}×64$$

$$=\dfrac {128\,\pi}{3}$$

### 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

$$\dfrac {512\,\pi}{7}$$

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 $$\rightarrow\;2\pi\,r\,h \;\Delta r$$

#### 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 $$0\leq a <b$$

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

Radius of cylindrical shell = $$2-x$$

Height of cylindrical shell = $$x^4$$

$$\therefore$$ Volume by cylindrical shells

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

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

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

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

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

$$=\dfrac {14\,\pi\,}{30}$$

$$=\dfrac {7\,\pi\,}{15}$$

### 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

# 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 $$\rightarrow\;2\pi\,r\,h \;\Delta r$$

#### 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)

$$\therefore$$ Volume by cylindrical shells

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

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

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

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

$$=\dfrac {62\,\pi\,}{5}$$

### 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 $$\rightarrow \;\;2\pi\,r\,h \;\Delta r$$

#### 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 $$0\leq a <b$$

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

Intersection point

$$x^2=6x-2x^2$$

$$3x^2-6x=0$$

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

$$x=0,\;x=2$$

$$\therefore$$ Volume by cylindrical shells

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

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

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

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

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

$$=8\,\pi\,$$

### 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