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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

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

 

Illustration Questions

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

Illustration Questions

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

Illustration Questions

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

Illustration Questions

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

Illustration Questions

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.

Illustration Questions

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

Illustration Questions

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

Illustration Questions

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

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