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### Area Of Surface Of Revolution

Learn calculation of surface area calculus of revolution & about y-axis, practice approximating the value of area of surface of revolution by Simpson's rule with using calculator.

# Area of Surface of Revolution

• A surface of revolution is formed when a curve is rotated about a line.
• We approximate the original curve by a polygon, when this polygon is rotated about an axis it creates simpler surface whose surface area approximates the actual surface area. By taking limits we can determine the exact surface area.
• The approximating surface has number of bands, each formed by rotating a line segment about an axis. To find the surface area each of these bands are considered a portion of circular cone.

• The area of band $$=A=\pi\, r_2(l_1+l)-\pi\,r_1l_1$$

$$=\pi[(r_2-r_1)l_1+r_2l]$$ ...(1)

Now from similar triangles ABC and ADE

$$\dfrac{l_1}{r_1}=\dfrac{l_1+l}{r_2}$$

$$\Rightarrow\,l_1r_2=l_1r_1+lr_1$$

Using this in (1)

$$\Rightarrow\,A=\pi(r_1l+r_2l)$$

$$=\pi\,l(r_1+r_2)$$

$$=2\pi rl$$

where  $$r=\dfrac{r_1+r_2}{2}$$ = average radius.

• Now consider the surface shown.

It is obtained by rotating by  $$y=f(x)$$,  $$a\leq x \leq b$$  about $$x$$ axis, where $$f$$ is positive and has continuous derivative. Divide (a, b) into $$n$$ parts each measuring  $$\dfrac{b-a}{x}=\Delta x$$. The part of the curve between $$x_{i-1}$$ and $$x_i$$ is approximated by line segment $$|P_{i-1}P_i|$$ and rotated about $$x$$ axis.

$$\therefore\,$$ Surface area of band is $$=2\pi \dfrac{(y_{i-1}+y_i)}{2}\,|P_{i-1}P_i|$$

$$\cong2\pi \dfrac{(y_{i-1}+y_i)}{2}×\sqrt{1+f'(x_1^*)^2}$$

$$\therefore$$ Area of revolution $$\cong\;\displaystyle\sum\limits^n_{i-1} 2\pi f(x_1^*)\sqrt{1+(f(x_1^*))^2}$$

as  $$n-\infty$$, we recognize this as Riemann Sum.

$$\Rightarrow$$ Area of surface of revolution  $$\displaystyle =\int \limits ^b_a2\pi f(x)\sqrt{1+(f'(x))^2}dx$$

In Leibniz notation, the formula becomes

$$\displaystyle S=\int\limits ^b_a\left(2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\right)dx$$

• If the curve is  $$x=g(y)$$, where  $$c\leq y\leq d$$ ,the formula for surface area becomes-

$$\displaystyle S=\int\limits ^d_c2\pi y\sqrt{1+\left(\dfrac{dx}{dy}\right)^2}dy$$

• If the rotation is done about  $$y$$ axis, then the area of surface of revolution of $$y=f(x)$$ ($$a\leq x\leq b$$ ) is given by

$$\displaystyle S=\int\limits ^b_a2\pi x\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx$$

• If the rotation of the curve $$x=f(y)$$ ( $$c\leq y\leq d$$ ) is done about  $$y$$ axis, then the area of surface of revolution is given by

$$\displaystyle S=\int\limits ^d_c2\pi x\sqrt{1+\left(\dfrac{dx}{dy}\right)}dy$$

#### Set up an integral for area of surface obtained by rotating the curve  $$y=\sqrt{1+4x}$$, $$1\leq x \leq5$$

A $$\displaystyle S=2\pi\int\limits ^5_1\sqrt{5+4x}\;dx$$

B $$\displaystyle S=2\pi\int\limits ^5_1\sqrt{1+4x}\;dx$$

C $$\displaystyle S=\pi\int\limits ^6_2\sqrt{x^2+5x+1}\;dx$$

D $$\displaystyle S=2\pi\int\limits ^7_2\sqrt{1+4x}\;dx$$

×

Area of surface of revolution obtained by rotating the curve $$y=f(x)$$ about $$x$$ axis when  $$a\leq x \leq b$$ , is

$$\displaystyle S=\int\limits ^b_a2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx$$

In this case $$a=1,\;b=5,\;f(x)=\sqrt{1+4x}=y$$

$$\dfrac{dy}{dx}=\dfrac{d}{dx}(\sqrt{1+4x})$$

$$=\dfrac{1×4}{2\sqrt{1+4x}}$$

$$=\dfrac{2}{(1+4x)^{1/2}}$$

$$\displaystyle \therefore \;S=\int\limits ^5_12\pi\sqrt{1+4x}×\sqrt{1+\dfrac{4}{1+4x}}dx$$

$$\displaystyle =\int \limits ^5_12\pi\;\sqrt{1+4x}\sqrt{\dfrac{5+4x}{1+4x}}dx$$

$$\displaystyle =2\pi\int\limits ^5_1\sqrt{5+4x}\;dx\to$$ which is the required integral.

### Set up an integral for area of surface obtained by rotating the curve  $$y=\sqrt{1+4x}$$, $$1\leq x \leq5$$

A

$$\displaystyle S=2\pi\int\limits ^5_1\sqrt{5+4x}\;dx$$

.

B

$$\displaystyle S=2\pi\int\limits ^5_1\sqrt{1+4x}\;dx$$

C

$$\displaystyle S=\pi\int\limits ^6_2\sqrt{x^2+5x+1}\;dx$$

D

$$\displaystyle S=2\pi\int\limits ^7_2\sqrt{1+4x}\;dx$$

Option A is Correct

#### Find the exact area of surface obtained by rotating the curve  $$y=x^3$$ about $$x$$ axis$$(0\leq x \leq2)$$

A 2.4236

B 517.1892

C 202.9407

D 5.7983

×

Area of surface of revolution obtained by rotating the curve $$y=f(x)$$, about $$x$$ axis when $$a\leq x \leq b$$, is

$$\displaystyle S=\int\limits ^b_a2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx$$

In this case $$a=0,\;b=2,\;y=f(x)=x^3$$

$$\dfrac{dy}{dx}=f'(x)=3x^2$$

$$\displaystyle \therefore \;S=\int\limits ^2_02\pi x^3\sqrt{1+9x^4}\;dx$$

$$\displaystyle =2\pi\int\limits ^2_0x^3\sqrt{1+9x^4}\;dx$$

put  $$1+9x^4=t$$

$$\Rightarrow36x^3\;dx=dt$$

$$\Rightarrow x^3\;dx=\dfrac{1}{36}\;dt$$

where $$x=2,\;t=145$$

$$x=0,\;t=1$$

$$\displaystyle \therefore \;S=2\pi\int\limits ^{145}_1\dfrac{1}{36}×\sqrt t\;dt$$

$$=2\pi \left[\dfrac{1}{36}\dfrac{t^{3/2}}{\dfrac{3}{2}}\right]^{145}_1$$

$$=\dfrac{\pi}{27}\left[(145)^{3/2}-1\right]$$

$$=202.9407$$

### Find the exact area of surface obtained by rotating the curve  $$y=x^3$$ about $$x$$ axis$$(0\leq x \leq2)$$

A

2.4236

.

B

517.1892

C

202.9407

D

5.7983

Option C is Correct

#### The curve $$x=\sqrt y$$  is rotated about  $$y$$ axis ($$1\leq y \leq 9$$), find the area of resulting surface.

A 203.68

B 111.93

C 5.24

D 0.02

×

The area of surface of revolution of  $$x=f(y)$$ $$(c\leq y\leq d)$$ about $$y$$ axis is given by

$$\displaystyle S=\int\limits ^d_c2\pi x\sqrt{1+\left(\dfrac{dx}{dy}\right)^2}dy$$

In this case $$c=1,\;d=9,\;x=f(y)=\sqrt y$$

$$\dfrac{dx}{dy}=\dfrac{1}{2\sqrt y}$$

$$\displaystyle \therefore \;S=\int\limits ^9_12\pi \sqrt y\sqrt{1+\dfrac{1}{4y}}\;dy$$

$$\displaystyle =2\pi\int \limits ^9_1\sqrt y\;\dfrac{\sqrt {4y+1}}{2\sqrt y}dy$$

$$\displaystyle =\pi\int \limits ^9_1\sqrt{1+4y}\;dy$$

$$=\pi\left[\dfrac{(1+4y)^{3/2}}{\dfrac{3}{2}×4}\right]^9_1$$

$$=\dfrac{\pi}{6}\left[(37)^{3/2}-5^{3/2}\right]$$

$$=\dfrac{\pi}{6}\left[225.06-11.18\right]$$

$$=\dfrac{\pi}{6}×213.88$$

$$=111.93$$

### The curve $$x=\sqrt y$$  is rotated about  $$y$$ axis ($$1\leq y \leq 9$$), find the area of resulting surface.

A

203.68

.

B

111.93

C

5.24

D

0.02

Option B is Correct

# Approximating the Value of Area of Surface of Revolution by Simpson's Rule

• If a curve $$y=f(x)$$ $$(a\leq x\leq b)$$ is rotating about $$x$$ axis then the area of surface revolution is given by

$$\displaystyle S=\int \limits^b_a2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\;dx$$

• If the curve is in the form $$x=f(y)$$ $$(c\leq y \leq d)$$ then the area of surface of revolution about $$x$$ axis is given by

$$\displaystyle S=\int \limits^d_c2\pi y\sqrt{1+\left(\dfrac{dx}{dy}\right)^2}\;dy$$

• If the curve $$y=f(x)$$ $$(a\leq x\leq b)$$ is rotated about $$y$$ axis then the area of surface of revolution is given by

$$\displaystyle S=\int \limits^b_a2\pi x\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\;dx$$

• If the curve $$x=f(y)$$ $$(c\leq y\leq d)$$ is rotated about $$y$$ axis then the area of surface of revolution is given by

$$\displaystyle S=\int \limits^d_c2\pi x\sqrt{1+\left(\dfrac{dx}{dy}\right)^2}\;dy$$

• Often the above integrals are not easy to evaluate because the indefinite integral cannot be expressed in terms of known function. Therefore, approximation method of integrals is used.

#### Use Simpson's Rule with $$n=8$$ to approximate the area of surface obtained by rotating the curve $$y=x+x^2$$ $$(0\leq x \leq 1)$$ about $$x$$ axis.

A 14.1522

B 5.6278

C 105.2125

D 78.3146

×

Area of surface of revolution obtained by rotating the curve $$y=f(x)$$ about  $$x$$ axis when $$a\leq x\leq b$$, is

$$\displaystyle S=\int \limits^b_a2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\;dx$$

In this case $$a=0,\;b=1,\,y=f(x)=x+x^2$$

$$\dfrac{dy}{dx}=f'(x)=1+2x$$

$$\displaystyle \therefore S=\int \limits^1_02\pi (x+x^2)\sqrt{1+(1+2x)^2}\;dx$$

$$\displaystyle =2\pi\int \limits^1_0 (x+x^2)\sqrt{4x^2+4x+2}\;dx$$

To evaluate the above integral we take $$n=8$$ in Simpson's Rule.

Simpson's Rule says that

$$\displaystyle \int \limits^b_af(x)dx\cong S_n=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+...............2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right]$$

where $$n$$ is even,  $$x_i=x_0+i\Delta x,\;\Delta x=\dfrac{b-a}{n}$$

$$\displaystyle\therefore \,S=2\pi\int \limits ^1_0(x+x^2)(\sqrt{4x^2+4x+2})\;dx\simeq S_8$$

$$=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+2f(x_6)+4f(x_7)+f(x_8)\right]$$

Here  $$\Delta x=\dfrac{1-0}{8}=\dfrac{1}{8},\;x_0=0,\;x_1=\dfrac{1}{8},\;x_2=\dfrac{1}{4},\;x_3=\dfrac{3}{8},\;x_4=\dfrac{1}{2},\;x_5=\dfrac{5}{8},\;x_6=\dfrac{6}{8},\;x_7=\dfrac{7}{8},\;x_8=b=1$$

$$f(x)=(x+x^2)\sqrt{4x^2+4x+2}$$

$$\therefore\,S\simeq 2\pi\left[\dfrac{1}{24}\;\left(f(0)+4f\left(\dfrac{1}{8}\right)+2f\left(\dfrac{1}{4} \right)+4f\left(\dfrac{3}{8}\right)+2f\left(\dfrac{1}{2}\right)+4f\left(\dfrac{5}{8}\right)+2f\left(\dfrac{6}{8}\right)+4f\left(\dfrac{7}{8}\right)+f(1)\right)\right]$$

$$=14.1522$$

### Use Simpson's Rule with $$n=8$$ to approximate the area of surface obtained by rotating the curve $$y=x+x^2$$ $$(0\leq x \leq 1)$$ about $$x$$ axis.

A

14.1522

.

B

5.6278

C

105.2125

D

78.3146

Option A is Correct

#### If the curve  $$x=1+2y^2$$  $$(1\leq y\leq 2)$$ is rotated about $$x$$ axis, find the area of surface of revolution.

A 59.39

B 673.24

C 109.12

D 5.81

×

If the curve  $$x=f(y)$$  $$(c\leq y\leq d)$$ is rotated about  $$x$$ axis then area = S

$$\displaystyle =\int\limits ^d_c2\pi y\sqrt{1+\left(\dfrac{dx}{dy}\right)^2}dy$$

In this case $$c=1,\;d=2,\;x=f(y)=1+2y^2$$

$$\dfrac{dx}{dy}=4y$$

$$\displaystyle \therefore \;S=\int\limits ^2_12\pi y\sqrt{1+16y^2}\;dy$$

$$\displaystyle =2\pi\int\limits ^2_1y\,\sqrt{1+16y^2}\;dy$$

put  $$1+16y^2=t$$

$$\Rightarrow\,32y\;dy=dt$$

$$\Rightarrow\,y\;dy=\dfrac{dt}{32}$$

when,  $$y=2,\;t=65$$

$$y=1,\;t=17$$

$$\displaystyle \therefore \;S=2\pi\int\limits ^{65}_{17}\dfrac{1}{32}×\sqrt t\;dt$$

$$=\left[\dfrac{2\pi}{32}\;\dfrac{t^{3/2}}{\dfrac{3}{2}}\right]^{65}_{17}$$

$$=\dfrac{\pi}{24}\left[(65)^{3/2}-(17)^{3/2}\right]$$

$$=\dfrac{\pi}{24}\left[524.05-70.09\right]$$

$$=59.39$$

### If the curve  $$x=1+2y^2$$  $$(1\leq y\leq 2)$$ is rotated about $$x$$ axis, find the area of surface of revolution.

A

59.39

.

B

673.24

C

109.12

D

5.81

Option A is Correct

#### The curve $$y=\sqrt{4-x^2}$$ ( $$0\leq x \leq\sqrt3$$ ) is rotated about  $$y$$ axis, find the area of resulting surface.

A $$25\pi$$

B $$4\pi$$

C $$562\pi$$

D $$\pi$$

×

The area of surface of revolution of  $$y=f(x)$$$$(a\leq x\leq b)$$ about $$y$$axis is given by

$$\displaystyle S=\int\limits ^b_a2\pi x\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx$$

In this case $$a=0,\;b=\sqrt3,\;f(x)=\sqrt{4-x^2}=y$$

$$f'(x)=\dfrac{dy}{dx}$$

$$=\dfrac{1×(-2x)}{2\sqrt{4-x^2}}$$

$$=\dfrac{-x}{\sqrt{4-x^2}}$$

$$\displaystyle \therefore \;S=\int\limits ^\sqrt3_02\pi x\sqrt{1+\dfrac{x^2}{4-x^2}}dx$$

$$\displaystyle =2\pi\int \limits ^\sqrt3_0x×\dfrac{2}{\sqrt{4-x^2}}$$

$$\displaystyle =4\pi\int \limits ^\sqrt3_0\dfrac{x}{\sqrt{4-x^2}}dx$$

Put  $$4-x^2=t$$

$$\Rightarrow-2xdx=dt$$

When, $$x=\sqrt3,\;t=1$$ and $$x=0,\;t=4$$

$$\displaystyle \therefore \;S=4\pi\int\limits ^1_4\dfrac{-1}{2}×\dfrac{1}{\sqrt t}dt$$

$$\displaystyle =2\pi\int \limits ^4_1\dfrac{1}{\sqrt{t}}dt$$

$$=2\pi\left[\dfrac{t^{1/2}}{\dfrac{1}{2}}\right]^4_1$$

$$=4\pi[\sqrt4-\sqrt1]$$

$$=4\pi×(2-1)$$

$$=4\pi$$

### The curve $$y=\sqrt{4-x^2}$$ ( $$0\leq x \leq\sqrt3$$ ) is rotated about  $$y$$ axis, find the area of resulting surface.

A

$$25\pi$$

.

B

$$4\pi$$

C

$$562\pi$$

D

$$\pi$$

Option B is Correct

#### Use calculator to evaluate the area of surface of revolution when $$y=e^{-x^2}$$, is rotated about $$x$$ axis $$(-1\leq x \leq1)$$

A 1.2346

B 596.1908

C 76.1029

D 11.0698

×

Area of surface of revolution obtained by rotating the curve $$y=f(x)$$ about $$x$$ axis when $$a\leq x \leq b$$ is

$$\displaystyle S=\int\limits ^b_a2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dy$$

In this case $$a=-1,\;b=1,\;f(x)=e^{-x^2}=y$$

$$\dfrac{dy}{dx}=e^{-x^2}×(-2x)$$

$$\displaystyle \therefore \;S=\int\limits ^1_{-1}2\pi e^{-x^2}\left(\sqrt{1+4x^2 e^{-2x^2}}\right)dx$$

$$\displaystyle =2\pi\int \limits ^1_{-1} e^{-x^2}\;\sqrt{1+4x^2e^{-2x^2}}dx$$

Now use calculator

$$=2\pi×1.7627$$

$$=11.0698$$

### Use calculator to evaluate the area of surface of revolution when $$y=e^{-x^2}$$, is rotated about $$x$$ axis $$(-1\leq x \leq1)$$

A

1.2346

.

B

596.1908

C

76.1029

D

11.0698

Option D is Correct

#### Use calculator to find the area of revolution when $$y=\dfrac{1}{4}x^2-\dfrac{lnx}{2}$$, $$1\leq x \leq 2$$ is rotated about $$y$$ axis.

A $$2\pi$$

B $$\dfrac{10\pi}{3}$$

C $$\dfrac{7\pi}{5}$$

D $$\dfrac{512\pi}{3}$$

×

The area of surface of revolution of $$y=f(x)$$$$a\leq x \leq b$$ about $$x$$ axis is given by

$$\displaystyle S=\int\limits ^b_a2\pi x\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx$$

In this case $$a=1,\;b=2,\;y=f(x)=\dfrac{x^2}{4}-\dfrac{lnx}{2}$$

$$\Rightarrow\dfrac{dy}{dx}=f'(x)=\dfrac{2x}{4}-\dfrac{1}{2x}$$

$$=\dfrac{x}{2}-\dfrac{1}{2x}$$

$$=\dfrac{1}{2}\left(x-\dfrac{1}{x}\right)$$

$$\displaystyle \therefore \;S=\int\limits ^2_12\pi x\sqrt{1+\left(\dfrac{1}{2}\left(x-\dfrac{1}{x}\right)\right)^2}dx$$

$$\displaystyle =2\pi\int \limits ^2_1x\;\sqrt{1+\dfrac{1}{4}\left(x^2+\dfrac{1}{x^2}-2\right)}dx$$

$$\displaystyle =\dfrac{2\pi}{2}\int \limits ^2_1x\;\sqrt{x^2+\dfrac{1}{x^2}+2}\;dx$$

$$\displaystyle =\pi\int \limits ^2_1x\;\sqrt{\left(x+\dfrac{1}{x}\right)^2}dx$$

$$\displaystyle =\pi\int \limits ^2_1x\;{\left(x+\dfrac{1}{x}\right)}dx$$

$$\displaystyle =\pi\int \limits ^2_1(x^2+1)dx$$

Use calculator to evaluate the integral.

$$\therefore\,S=\pi×\dfrac{10}{3}$$

$$=10.4667$$

### Use calculator to find the area of revolution when $$y=\dfrac{1}{4}x^2-\dfrac{lnx}{2}$$, $$1\leq x \leq 2$$ is rotated about $$y$$ axis.

A

$$2\pi$$

.

B

$$\dfrac{10\pi}{3}$$

C

$$\dfrac{7\pi}{5}$$

D

$$\dfrac{512\pi}{3}$$

Option B is Correct