Finding the area between curves

[FARADAY JR]

Homework Statement

find the area bounded by :
4x+y^2=12
x=y

Homework Equations

The Attempt at a Solution

f(x)= √4x-12
g(X)= x
√4x-12=x
√3x-12=0I'm lost how am i supposed to set them equal to each other if the second equation is y=x? can anybody help me please

 

[tiny-tim]

Hi FARADAY JR!

(pleeease don't give different threads the same name! )

find the area bounded by :
4x+y^2=12
x=y
…
I'm lost how am i supposed to set them equal to each other if the second equation is y=x?

(where did √3x-12=0 come from? )

just replace y² by x²

 

[FARADAY JR]

Hi FARADAY JR!

(pleeease don't give different threads the same name! )


(where did √3x-12=0 come from? )

just replace y² by x²

Ok, So
4x+x^2-12=x
3x+x^2-12=0
3 (x+2)(x-2)=0 ?

 

[FARADAY JR]

what do i do next?

 

[tiny-tim]

4x+x^2-12=x

nooo … where did that x on the RHS come from?

 

[FARADAY JR]

nooo … where did that x on the RHS come from?

I was trying to do f(x)-g(x)=0
so I can find the high and low

 

[FARADAY JR]

so should I do it like this:
4x+x^2-12=0
(x+6)(x-2)=0
where X = 2,-6
?

 

[tiny-tim]

Yup!

(and now can you get the area? )