# Surface area limited with two functions

#### wishmaster

##### Active member
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

Staff member
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
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

Staff member
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
Correct. Have you determined which function is the "top" function on the relevant interval?
its $g(x)$

#### MarkFL

Staff member
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
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

Staff member
Its $h(x)$, im sorry.
Correct. Now, can you write the definite integral representing the area bounded by the two functions?

#### wishmaster

##### Active member
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

Staff member
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
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

Staff member
i will need some time to figure this out.....
Here is a plot of the area to be found:

Can you see the symmetry now?

#### MarkFL

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