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Surface area of rotation about the y-axis

Petrus

Well-known member
Feb 21, 2013
739
Calculate the area of ​​the surface of rotation which occurs when the curve
rotate in y-axe.
I start with \(\displaystyle x=\sqrt{28y}\) then \(\displaystyle f'(x)=\frac{14}{\sqrt{28y}}\)
so we got
\(\displaystyle 2\pi\int_0^{21}\sqrt{28y}\sqrt{1+(\frac{14}{\sqrt{28y}})}^2\)
then I rewrite as \(\displaystyle \int_0^2\sqrt{28y}\sqrt{1+\frac{196}{28y}}\)
\(\displaystyle \sqrt{28y}\sqrt{1+\frac{196}{28y}}<=>\sqrt{28y+196}\)
So I got \(\displaystyle 2\pi\int_0^{21}\sqrt{28y+196}\) and if Integrate it I get
\(\displaystyle \frac{(28y+196)^{1.5}}{1.5*28}\) Is this correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You have the right idea, but you want to write things in terms of $x$.

\(\displaystyle S=2\pi\int_a^b x\sqrt{1+\left[f'(x) \right]^2}\,dx\)

You want the lower limit of integration to be the $x$-value that corresponds with $y=0$, and the upper limit to correspond with $y=21$.

If \(\displaystyle f(x)=\frac{x^2}{28}\), then what is $f'(x)$?

edit: I see now that your approach is valid, and easier to integrate. (Smile)

Your anti-derivative is correct (except don't forget the $2\pi$), so apply the FTOC to find the value of the definite integral.
 
Last edited:

Petrus

Well-known member
Feb 21, 2013
739
You have the right idea, but you want to write things in terms of $x$.

\(\displaystyle S=2\pi\int_a^b x\sqrt{1+\left[f'(x) \right]^2}\,dx\)

You want the lower limit of integration to be the $x$-value that corresponds with $y=0$, and the upper limit to correspond with $y=21$.

If \(\displaystyle f(x)=\frac{x^2}{28}\), then what is $f'(x)$?

edit: I see now that your approach is valid, and easier to integrate. (Smile)
\(\displaystyle \frac{x^3}{84}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \frac{x^3}{84}\)
You integrated, rather than differentiated, but disregard what I posted about that, your method is correct and easier to use.
 

Petrus

Well-known member
Feb 21, 2013
739
You integrated, rather than differentiated, but disregard what I posted about that, your method is correct and easier to use.
So the answer is \(\displaystyle 2\pi*\frac{(28*21+196)^{1.5}}{1.5*28}\) I am correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
No, you need to evaluate:

\(\displaystyle 2\pi\left[\frac{(28y+196)^{\frac{3}{2}}}{42} \right]_0^{21}=\frac{\pi}{21}\left((28\cdot21+196)^{\frac{3}{2}}-(28\cdot0+196)^{\frac{3}{2}} \right)=?\)
 

Petrus

Well-known member
Feb 21, 2013
739
No, you need to evaluate:

\(\displaystyle 2\pi\left[\frac{(28y+196)^{\frac{3}{2}}}{42} \right]_0^{21}=\frac{\pi}{21}\left((28\cdot21+196)^{\frac{3}{2}}-(28\cdot0+196)^{\frac{3}{2}} \right)=?\)
\(\displaystyle \frac{\pi}{21}((28*21+196)^{\frac{3}{2}}-196^{\frac{3}{2}})\) I don't wanna use any calculator so that would be my answer.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Without using a calculator, you could find:

\(\displaystyle 28\cdot21+196=4(3\cdot7^2+49)=16\cdot49=28^2\)

and so:

\(\displaystyle (28^2)^{\frac{3}{2}}=28^3\)

and then:

\(\displaystyle 28\cdot0+196=14^2\)

and then you have:

\(\displaystyle S=\frac{\pi}{21}(28^3-14^3)=\frac{14^3\pi}{21}(2^3-1)=\frac{14^3\pi}{3}=\frac{2744\pi}{3}\)

all easily done without a calculator.
 

Petrus

Well-known member
Feb 21, 2013
739
Without using a calculator, you could find:

\(\displaystyle 28\cdot21+196=4(3\cdot7^2+49)=16\cdot49=28^2\)

and so:

\(\displaystyle (28^2)^{\frac{3}{2}}=28^3\)

and then:

\(\displaystyle 28\cdot0+196=14^2\)

and then you have:

\(\displaystyle S=\frac{\pi}{21}(28^3-14^3)=\frac{14^3\pi}{21}(2^3-1)=\frac{14^3\pi}{3}=\frac{2744\pi}{3}\)

all easily done without a calculator.
hmm I get wrong with FTOC
If we subsitute \(\displaystyle u=28y+196\) then we get\(\displaystyle \frac{\pi}{4}\int_{196}^{784}\sqrt{28y+196}\)

edit: I am confused with this change of 2pi. Some does subsitute and then change that 2pi and other does not. I have been serching over internet and don't understand that.
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
hmm I get wrong with FTOC
If we subsitute \(\displaystyle u=28y+196\) then we get\(\displaystyle \frac{\pi}{4}\int_{196}^{784}\sqrt{28y+196}\)

edit: I am confused with this change of 2pi. Some does subsitute and then change that 2pi and other does not. I have been serching over internet and don't understand that.
Using the substitution you cite, \(\displaystyle u=28y+196\), we get:

\(\displaystyle S=\frac{2\pi}{28}\int_{28\cdot0+196}^{28\cdot21+196}u^{\frac{1}{2}}\,du\)

\(\displaystyle S=\frac{\pi}{14}\int_{196}^{784}u^{\frac{1}{2}}\,du\)

\(\displaystyle S=\frac{\pi}{21}\left[u^{\frac{3}{2}} \right]_{196}^{784}=\frac{\pi}{21}(28^3-14^3)=\frac{2744\pi}{3}\)
 

Petrus

Well-known member
Feb 21, 2013
739
Using the substitution you cite, \(\displaystyle u=28y+196\), we get:

\(\displaystyle S=\frac{2\pi}{28}\int_{28\cdot0+196}^{28\cdot21+196}u^{\frac{1}{2}}\,du\)

\(\displaystyle S=\frac{\pi}{14}\int_{196}^{784}u^{\frac{1}{2}}\,du\)

\(\displaystyle S=\frac{\pi}{21}\left[u^{\frac{3}{2}} \right]_{196}^{784}=\frac{\pi}{21}(28^3-14^3)=\frac{2744\pi}{3}\)
Thanks again Mark:) Now I understand!:) Thanks for the \(\displaystyle \LaTeX\) code as well