Finding area between circle and y-axis using integration

[roger]

hi, I am stuck on this question for homework :

I'm not sure how to find the answer, as I thought integration is used only if the curve intersects at the x-axis ?


Thanx for any help


1.) The equation of circle x^-8x+y^2-10y=-16

It intersects the y-axis at two points : (0,2) and (0,8)

Find the small area between the curve and the y axis.(on the left of the y axis)


2.) Also, the chord between 0,2 and 0,8 on the circle subtends an angle at the centre (4,5).

Show that the angle is 7/25 using the cosine rule.


I'm not sure what to do here as well..

 

[dextercioby]

Plot that circle and then decide whether u'll be needing calculus to find that area.

As for the second,i don't follow.Is it the same circle...?

Daniel.

 

[roger]

Plot that circle and then decide whether u'll be needing calculus to find that area.

As for the second,i don't follow.Is it the same circle...?

Daniel.

The area is bound between the y-axis and the circle. But I don't see how else it could be done without calculus ?

The second question is the same circle and I need to find the angle .
(0,2) and (0,8) is the chord on the circle .

 

[dextercioby]

Alright,then,write the equation and see which branch of the explicitation will u be needing (hint:u can explicitate "x" as a function of "y",because you're given the intercepts with the Oy axis).

Daniel.

 

[roger]

Alright,then,write the equation and see which branch of the explicitation will u be needing (hint:u can explicitate "x" as a function of "y",because you're given the intercepts with the Oy axis).

Daniel.

I've not heard that word before...explicitate ..

but I don't understand what your saying.

 

[dextercioby]

Hmmm,alright.Your equation's typically

$$(x-x_{0})^{2}+(y-y_{0})^{2}=1$$

Express x=x(y).

Daniel.

 

[roger]

Hmmm,alright.Your equation's typically

$$(x-x_{0})^{2}+(y-y_{0})^{2}=1$$

Express x=x(y).

Daniel.

I don't get it


especially
Express x=x(y).

 

[dextercioby]

You need to put somethting,a function,in this integral

$$\mbox{Area}=\int_{\mbox{intercept no.1}}^{\mbox{intercept no.2}} f(y) \ dy$$

And that function of "y" (f(y)) is the one u'll be getting from the circle's equation.

Daniel.

 

[roger]

You need to put somethting,a function,in this integral

$$\mbox{Area}=\int_{\mbox{intercept no.1}}^{\mbox{intercept no.2}} f(y) \ dy$$

And that function of "y" (f(y)) is the one u'll be getting from the circle's equation.

Daniel.

okay, and then take away the the intercept 2 - 1 ?

but what does x(y)=x mean ? I need a thorough explanation on this please.

 

[dextercioby]

It's a notation;x=x(y) means the function "x" written in terms of the variable "y".

Daniel.

 

[roger]

It's a notation;x=x(y) means the function "x" written in terms of the variable "y".

Daniel.

but if i took away the 1st from the second integral is the order important ?

what difference does it make ?

If the area was not bound by either x or y-axis is there a direct way to calulate the area?

 

[dextercioby]

Yes,of course.The area of a plane domain $D\subseteq\mathbb{R}^{2}$ is given by

$$S_{D}=:\iint_{D} dx \ dy$$

as long as u choose Oxy as a set of orthormal coordinates in the plane.

Daniel.

 

[roger]

It's a notation;x=x(y) means the function "x" written in terms of the variable "y".

Daniel.

Daniel, what difference would it make if I wrote g(y)=x ?

 

[arildno]

Daniel, what difference would it make if I wrote g(y)=x ?

The difference is that x=x(y) is a "loose/sloppy" notation, whereas x=g(y) is an example of a "rigorous/careful" notation. That's basically it.