Find the area of a segment of a circle using integration

[nafisanazlee]

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?

 

[Mark44]

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.

 

[nafisanazlee]

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?

.

 

[PeroK]

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?

 

[nafisanazlee]

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.

 

[PeroK]

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.

 

[Mark44]

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.

 

[nafisanazlee]

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.

 

[vela]

You could also solve for $x$ as a function of $y$ and integrate with respect to $y$. That's pretty straightforward as well.