#1

Thanks for reading.

An equation for a circle is given, x^2-4x+y^2+6y=12

I've converted this to the standard circle equation:

(x-2)^2+(y+3)^2=25

It runs throught the x axis in (6;0) og (-2;0)

Now, I have to find the area of the circle that is ABOVE the x axis. I have no idea what to do 0.o

Any help is appreciated!

An equation for a circle is given, x^2-4x+y^2+6y=12

I've converted this to the standard circle equation:

(x-2)^2+(y+3)^2=25

It runs throught the x axis in (6;0) og (-2;0)

Now, I have to find the area of the circle that is ABOVE the x axis. I have no idea what to do 0.o

Any help is appreciated!

#2

Look in the back of the book.

#3

One, I would like to understand the process.

Two, there is no book -_- it's a copy sheet.

Two, there is no book -_- it's a copy sheet.

#4

Get the book.

#5

i can tell that radius is 5 and the center is (2 , -3) .... thats about it lol

#6

Draw a diagram.

Intergrate.

Use logic to find out between what limits to intergrate.

Intergrate.

Use logic to find out between what limits to intergrate.

#7

Lol, freshman geometry. This problem is so simple.

/doesn't remember

/doesn't remember

#8

the answer is 3.because when you add 1+2 you get three of course

but uhh sorry algebra 2 doesnt help me here

but uhh sorry algebra 2 doesnt help me here

#9

If you don't have to give any specifics EMGs_rule up there ^ got you the proper answer

#10

Draw a diagram.

Intergrate.

Use logic to find out between what limits to intergrate.

i just sat my maths mock exam and i never thought of that

integration is definately the way to go, you'll have to find where it cuts the x-axis first

#11

We have to find the exact area above the x-axis.

A friend of mine actually figured it out just now, you find the area of the slice of the circle from the middle of the circle out to the intersections with the x-axis, and then subtract the area of the triangle formed by the centrum to the intersections with the x-axis.

Thanks for trying, everyone :-)

A friend of mine actually figured it out just now, you find the area of the slice of the circle from the middle of the circle out to the intersections with the x-axis, and then subtract the area of the triangle formed by the centrum to the intersections with the x-axis.

Thanks for trying, everyone :-)

#12

Ok. First you have to calculate the area of the sector between the radii that start on the x axis. Then, you calculate the area of the triangle formed by the radii and the x axis. Then you subtract the second thing from the first thing, and then you have the area of the part under the x axis. Now you have to calculate the area of the entire circle, and then subtract from that the area of the under-part.

Voila!

I got beaten again...

Voila!

We have to find the exact area above the x-axis.

A friend of mine actually figured it out just now, you find the area of the slice of the circle from the middle of the circle out to the intersections with the x-axis, and then subtract the area of the triangle formed by the centrum to the intersections with the x-axis.

Thanks for trying, everyone :-)

I got beaten again...

#13

Ok. First you have to calculate the area of the sector between the radii that start on the x axis. Then, you calculate the area of the triangle formed by the radii and the x axis. Then you subtract the second thing from the first thing, and then you have the area of the part under the x axis. Now you have to calculate the area of the entire circle, and then subtract from that the area of the under-part.

Voila!

I got beaten again...

Still nicely done, and put

#14

^^^^

someones gettin an A lol

someones gettin an A lol

#15

Still it's mathematically incorrect cause if you use triangles, you're making small mistakes. I think if you want to be 100% accurate you need to calculate the function of curve above x axis (which is a part of the circle) and than calculate an integral from it.

*Last edited by dilbert_5150 at Feb 3, 2008,*

#16

Still it's mathematically incorrect cause if you use triangles, you're making small mistakes. I think if you want to be 100% accurate you need to calculate the function of curve above x axis (which is a part of the circle) and than calculate an integer from it.

Everything we stated could be done with formulas and equations, so it is mathematically precise. Your method might be valid, but I don't really get what you mean, tbh. GOt an example of the formula you'd use?

I have the answer, so if any nerds wanna see if they can do it, I can tell you if you're right Yes, I know I'm a nerd myself..

#17

Still it's mathematically incorrect cause if you use triangles, you're making small mistakes. I think if you want to be 100% accurate you need to calculate the function of curve above x axis (which is a part of the circle) and than calculate an integer from it.

I don't think you understand how exactly the solution works. I guarantee that there are no mistakes or approximations at all.

#18

The answer is (100.'pi' - 36) / 3 , right?

#19

you're right guys. It's just my poor English skills I haven't understood properly what you meant. It's pretty simple task, and can be calculated without any "high-end" methods. You came out with a clever solution. Anyway, trying to find out a integral for that task, could be cool ^^ (yeah, only for nerds...)

cheers

cheers

*Last edited by dilbert_5150 at Feb 3, 2008,*

#20

The answer is (100.'pi' - 36) / 3 , right?

Give or take 80.. It turns out to be 11,2.

#21

Give or take 80.. It turns out to be 11,2.

alright, I made a mistake. I corrected it but can't get to 11,2.

the arch above the x axis is 120º, the total area of the circle is 25'pi', and the area of the triangle you have to subtract from the section you want is 12 in your calculations?

#22

Well, you can make y=0 to find the x-intercept

(x-2)^2+(0+3)^2=25

(x-2)^2=16

x-2=4

x=6

then integrate the equation x^2-4x+y^2+6y=12

dang, i've completely forgotten implicit integration.

(x-2)^2+(0+3)^2=25

(x-2)^2=16

x-2=4

x=6

then integrate the equation x^2-4x+y^2+6y=12

dang, i've completely forgotten implicit integration.