MHB Surface area limited with two functions

[theakdad]

I have to calculate the surface area limited with two functions:

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

i was thinking first to calculate the points where this functions are intersecting by $$g(x)=h(x)$$ 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]

I have to calculate the surface area limited with two functions:

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

i was thinking first to calculate the points where this functions are intersecting by $$g(x)=h(x)$$ 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?

 

[theakdad]

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: $$-\frac{1}{2}, \frac{1}{2}$$

 

[MarkFL]

So intersections are: $$-\frac{1}{2}, \frac{1}{2}$$

Correct. Have you determined which function is the "top" function on the relevant interval?

 

[theakdad]

Correct. Have you determined which function is the "top" function on the relevant interval?

its $g(x)$

 

[MarkFL]

its $g(x)$

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

Which function is greater at $x=0$?

 

[theakdad]

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

Which function is greater at $x=0$?

Its $h(x)$, I am sorry.

 

[MarkFL]

Its $h(x)$, I am sorry.

Correct. Now, can you write the definite integral representing the area bounded by the two functions?

 

[theakdad]

Correct. Now, can you write the definite integral representing the area bounded by the two functions?

$$S=\int_a^b(h(x)+C)-\int_a^b(g(x)+C)$$ where $a$ is $$-\frac{1}{2}$$ and $b$ is $$\frac{1}{2}$$ ??

 

[MarkFL]

No, you want:

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

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:

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

 

[theakdad]

No, you want:

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

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:

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

i will need some time to figure this out...

 

[MarkFL]

i will need some time to figure this out...

Here is a plot of the area to be found:

View attachment 1722

Can you see the symmetry now?

 

[theakdad]

Here is a plot of the area to be found:

View attachment 1722

Can you see the symmetry now?

I see,yes. Whats the software you use to write graphs?

 

[MarkFL]

I see,yes. Whats the software you use to write graphs?

This is a link to the site is used, with the command I used to generate the graph:

y=4/(4x^2+1),y=2 where x=-1/2 to 1/2,y=0 to 4 - Wolfram|Alpha

 

[HallsofIvy]

$$S=\int_a^b(h(x)+C)-\int_a^b(g(x)+C)$$ where $a$ is $$-\frac{1}{2}$$ and $b$ is $$\frac{1}{2}$$ ??

You haven't yet done the integration so there should be no "C".