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Surface area limited with two functions

wishmaster

Active member
Oct 11, 2013
211
I have to calculate the surface area limited with two functions:

\(\displaystyle g(x)=2\) and \(\displaystyle h(x)=\frac{4}{4x^2+1}\)

i was thinking first to calculate the points where this functions are intersecting by \(\displaystyle g(x)=h(x)\) and to find corresponding $x$ for the intersection...and then to calculate definite integral somehow. Am i on the right path? Thank you for your help!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I have to calculate the surface area limited with two functions:

\(\displaystyle g(x)=2\) and \(\displaystyle h(x)=\frac{4}{4x^2+1}\)

i was thinking first to calculate the points where this functions are intersecting by \(\displaystyle g(x)=h(x)\) and to find corresponding $x$ for the intersection...and then to calculate definite integral somehow. Am i on the right path? Thank you for your help!
Yes, you are on the right path.

You may also utilize the even symmetry of the bounding functions to find only the first quadrant area, and then double that to get the total area.

Have you found the first quadrant point of intersection?
 

wishmaster

Active member
Oct 11, 2013
211
Yes, you are on the right path.

You may also utilize the even symmetry of the bounding functions to find only the first quadrant area, and then double that to get the total area.

Have you found the first quadrant point of intersection?
So intersections are: \(\displaystyle -\frac{1}{2}, \frac{1}{2}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
So intersections are: \(\displaystyle -\frac{1}{2}, \frac{1}{2}\)
Correct. Have you determined which function is the "top" function on the relevant interval?
 

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
its $g(x)$
Pick a test value for $x$ in the interval \(\displaystyle \left(-\frac{1}{2},\frac{1}{2} \right)\). I would suggest $x=0$ for simplicity.

Which function is greater at $x=0$?
 

wishmaster

Active member
Oct 11, 2013
211
Pick a test value for $x$ in the interval \(\displaystyle \left(-\frac{1}{2},\frac{1}{2} \right)\). I would suggest $x=0$ for simplicity.

Which function is greater at $x=0$?
Its $h(x)$, im sorry.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
Correct. Now, can you write the definite integral representing the area bounded by the two functions?
\(\displaystyle S=\int_a^b(h(x)+C)-\int_a^b(g(x)+C)\) where $a$ is \(\displaystyle -\frac{1}{2}\) and $b$ is \(\displaystyle \frac{1}{2}\) ??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
No, you want:

\(\displaystyle S=\int_{-\frac{1}{2}}^{\frac{1}{2}}\left[h(x)-g(x) \right]\,dx\)

Since the integrand is even and the limits to integration also symmetric about the $y$-axis, we may use the even-function rule to state:

\(\displaystyle S=2\int_{0}^{\frac{1}{2}}\left[h(x)-g(x) \right]\,dx\)
 

wishmaster

Active member
Oct 11, 2013
211
No, you want:

\(\displaystyle S=\int_{-\frac{1}{2}}^{\frac{1}{2}}\left[h(x)-g(x) \right]\,dx\)

Since the integrand is even and the limits to integration also symmetric about the $y$-axis, we may use the even-function rule to state:

\(\displaystyle S=2\int_{0}^{\frac{1}{2}}\left[h(x)-g(x) \right]\,dx\)
i will need some time to figure this out.....
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
\(\displaystyle S=\int_a^b(h(x)+C)-\int_a^b(g(x)+C)\) where $a$ is \(\displaystyle -\frac{1}{2}\) and $b$ is \(\displaystyle \frac{1}{2}\) ??
You haven't yet done the integration so there should be no "C".