Find the area of a segment of a circle using integration

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Mentor note: Moved from a math technical section, so template is not present.
I was asked to calculate the area of the smaller section enclosed by the circle x²+y²-6x-8y-35=0 and the x axis. I've tried to solve it with geometry, using the x-intercepts and the centre of the circle I drew a triangle. Then by subtracting the area of triangle from the area of one particular sector I got the answer. But I couldn't do it with calculus, using the actual function. Can you help me with it? How do I approach?
 
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nafisanazlee said:
I was asked to calculate the area of the smaller section enclosed by the circle x²+y²-6x-8y-35=0 and the x axis. I've tried to solve it with geometry, using the x-intercepts and the centre of the circle I drew a triangle. Then by subtracting the area of triangle from the area of one particular sector I got the answer. But I couldn't do it with calculus, using the actual function. Can you help me with it? How do I approach?
Complete the square in the x and y terms, and then isolate the factored y terms on one side. If you solve for y you will get two functions. One of the functions represents the y values above the x-axis and the other represents the y values below the x-axis. It's the latter of these that you want to use in the integral.

BTW, questions like this one should be posted in the Calculus & Beyond homework section, using the template. I'm going to move this thread to that section.
 
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Mark44 said:
Complete the square in the x and y terms, and then isolate the factored y terms on one side. If you solve for y you will get two functions. One of the functions represents the y values above the x-axis and the other represents the y values below the x-axis. It's the latter of these that you want to use in the integral.

BTW, questions like this one should be posted in the Calculus & Beyond homework section, using the template. I'm going to move this thread to that section.
Thanks for your reply. I actually tried this method but unfortunately this gives me two functions which are the upper half and the lower half of the circle. I get y = 4±√(-x²+6x+51). Could you tell me where I was making a mistake?

.
 

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nafisanazlee said:
Thanks for your reply. I actually tried this method but unfortunately this gives me two functions which are the upper half and the lower half of the circle. I get y = 4±√(-x²+6x+51). Could you tell me where I was making a mistake?

.
That all looks good to me. Why do you think there's a mistake?
 
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PeroK said:
That all looks good to me. Why do you think there's a mistake?
Mark said- "One of the functions represents the y values above the x-axis and the other represents the y values below the x-axis. It's the latter of these that you want to use in the integral." But the latter one also contains some positive values of y. Anyways, I now understand using the latter one will give me the area. I was just a bit confused. Thanks for the clarification.
 
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nafisanazlee said:
Mark said- "One of the functions represents the y values above the x-axis and the other represents the y values below the x-axis. It's the latter of these that you want to use in the integral." But the latter one also contains some positive values of y. Anyways, I now understand using the latter one will give me the area. I was just a bit confused. Thanks for the clarification.
I would calculaute the area below the x-axis (as this is a single function) and then subtract that from the total area of a circle to get the area above the x-axis.
 
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nafisanazlee said:
Mark said- "One of the functions represents the y values above the x-axis and the other represents the y values below the x-axis...."
I erred in saying the above. One of the functions, the one that starts with "4 + ..." represents the part of the circle above the line y = 4. The other represents the part below that line. To get the area of the part of the circle below the x axis, integrate, with respect to x, between the two x-intercepts. Make sure you set up your integral correctly or you could get a negative value.
 
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Mark44 said:
I erred in saying the above. One of the functions, the one that starts with "4 + ..." represents the part of the circle above the line y = 4. The other represents the part below that line. To get the area of the part of the circle below the x axis, integrate, with respect to x, between the two x-intercepts. Make sure you set up your integral correctly or you could get a negative value.
Thank you so much.
 
  • #9
You could also solve for ##x## as a function of ##y## and integrate with respect to ##y##. That's pretty straightforward as well.
 

Related to Find the area of a segment of a circle using integration

1. What is a segment of a circle?

A segment of a circle is the region enclosed by a chord and the corresponding arc lying between the chord's endpoints. It is essentially a "slice" of the circle that does not necessarily pass through the center.

2. How do you set up the integral to find the area of a segment of a circle?

To set up the integral, you need to define the circle's equation, typically \(x^2 + y^2 = r^2\). Then, determine the limits of integration based on the chord's endpoints. The area of the segment can be found by integrating the function that describes the upper semicircle and subtracting the area under the chord (a straight line).

3. What is the integral formula to find the area of a segment of a circle?

The integral formula to find the area of a segment of a circle is:\[ A = \int_{a}^{b} \left( \sqrt{r^2 - x^2} - y_{\text{chord}} \right) \, dx \]where \( \sqrt{r^2 - x^2} \) represents the upper semicircle and \( y_{\text{chord}} \) is the linear equation of the chord.

4. How do you find the equation of the chord for the integration?

The equation of the chord can be found using the endpoints of the chord. If the endpoints are \((x_1, y_1)\) and \((x_2, y_2)\), the slope \(m\) of the chord is \(\frac{y_2 - y_1}{x_2 - x_1}\). The equation of the chord is then \( y = mx + c \), where \(c\) can be found using one of the endpoints to solve for the y-intercept.

5. Can you provide an example of calculating the area of a segment of a circle using integration?

Sure! Consider a circle with radius \(r = 5\) and a chord from \((-3, 4)\) to \((3, 4)\). The upper semicircle is described by \( y = \sqrt{25 - x^2} \). The chord is a horizontal line at \( y = 4 \). The area of the segment is:\[ A = \int_{-3}^{3} \left( \sqrt{25 - x^2} - 4 \right) \, dx \]Evaluating this integral will give the area of the segment.

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