MHB Area of 'that' part of the circle ....

[marutkpadhy]

Using integral find the area of that part of the circle x^2 + y^2 = 16 which is exterior to the parabola y^2 = 6x.

 

[MarkFL]

I would use the following diagram:

View attachment 2707

We know half of the circle is exterior to the parabola, which is the part to the left of the $y$-axis. We know both curves are symmetrical about the $x$-axis, so we need only find the area in green, double it, then add it to the half-circle.

Can you set up the integral to find the area in green?

 

[marutkpadhy]

No :/ I am very new to this topic, I am learning. I would appreciate if you could give the detailed solution, please. :)

Also, I am very confused with this LaTeX system, and expressing mathematics on web. :(

 

[MarkFL]

Providing a detailed solution won't help you nearly as much as guiding you through the problem, so that you are an active participant in the process.

First, let's determine our limits of integration. The left limit is easy, it is just $x=0$. For the right limit, we need to determine where the circle and parabola intersect. So, solve both equations for $y^2$, and equate the two results, and then solve for $x$, discarding any roots that do not apply. What do you find?

 

[marutkpadhy]

x = 2 or -8, discarding -8, we have only solution for x = 2.
now?

 

[MarkFL]

x = 2 or -8, discarding -8, we have only solution for x = 2.
now?

Yes, good. So, how do we find the area between two curves on a given interval?

 

[marutkpadhy]

By Integrating,
f(x) - g(x)
where these two functions are of the curves.
Now? Please guide me the whole solution.

 

[Prove It]

By Integrating,
f(x) - g(x)
where these two functions are of the curves.
Now? Please guide me the whole solution.

Why? You already have said what it is you have to do. Write down what f(x) - g(x) is and integrate it over the given region...

 

[MarkFL]

By Integrating,
f(x) - g(x)
where these two functions are of the curves.
Now? Please guide me the whole solution.

Correct...can you write the integral? You know the limits of integration, and you simply need to solve for $y$ in the two curves, taking the first quadrant roots. What is the integral representing the area I shaded above in green?

 

[marutkpadhy]

Integrating ( \sqrt{16-x^2} - \sqrt{6x} ),
lower limit = 0,
upper limit = 2
right?
then doubling the answer,
+ \pi r^2 / 2 = 8

 

[MarkFL]

Yes, you have the right idea (I don't know how you got an answer of 8 though). The total area $A$ is:

$$A=\frac{1}{2}\pi\cdot4^2+2\int_0^2 \sqrt{16-x^2}-\sqrt{6x}\,dx$$

The second term in the integrand is a straightforward application of the power rule, but can you tell me what type of substitution would be appropriate for the first term?