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

  • MHB
  • Thread starter marutkpadhy
  • Start date
  • Tags
    Area Circle
In summary: The first term in the integrand is a straightforward application of the power rule, so a power series substitution would be appropriate.
  • #1
marutkpadhy
9
0
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.
 
Physics news on Phys.org
  • #2
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?
 

Attachments

  • circleparabola.jpg
    circleparabola.jpg
    17.4 KB · Views: 70
  • #3
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. :(
 
  • #4
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?
 
  • #5
x = 2 or -8, discarding -8, we have only solution for x = 2.
now?
 
  • #6
marutkpadhy said:
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?
 
  • #7
By Integrating,
f(x) - g(x)
where these two functions are of the curves.
Now? Please guide me the whole solution.
 
  • #8
marutkpadhy said:
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...
 
  • #9
marutkpadhy said:
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?
 
  • #10
Integrating ( \sqrt{16-x^2} - \sqrt{6x} ),
lower limit = 0,
upper limit = 2
right?
then doubling the answer,
+ \pi r^2 / 2 = 8
 
  • #11
Yes, you have the right idea (I don't know how you got an answer of 8 though). The total area $A$ is:

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

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?
 

FAQ: Area of 'that' part of the circle ....

What is the formula for finding the area of a circle?

The formula for finding the area of a circle is A = πr², where A is the area and r is the radius of the circle.

How do you find the area of a sector of a circle?

The formula for finding the area of a sector of a circle is A = (θ/360)πr², where θ is the central angle of the sector and r is the radius of the circle.

Can the area of a circle be negative?

No, the area of a circle cannot be negative as it represents the amount of space enclosed by the circle and must always be a positive value.

What units are used to measure the area of a circle?

The area of a circle is typically measured in square units such as square inches, square feet, or square meters.

How does the area of a circle change if the radius is doubled?

If the radius of a circle is doubled, the area of the circle will increase by a factor of 4. This is because the area is directly proportional to the square of the radius.

Back
Top