I'm trying to use Integrals, what do you think?

 

But this exercice is from Vectors and Analitic Geometry PDF, do you have another way of solving it?

And I just can not find the center... can you help me?

Try a linear substitution so 3x' = 3x + a and choose "a" so that the 72x term disappears. Then do the same for y

First, "9 x^2 + 9y^2 + 72 x − 12 y + 103 = 0" is the equation of a curve in the xy-plane and nether has an "area". I presume you mean "find the area of the disk bounded by the circle described by 9 x^2 + 9y^2 + 72 x − 12 y + 103 = 0".

 

The first thing I would do is complete the squares in both x and y to write this in "standard form":

9(x^2- 8x+ 16)+ 9(y^2- (4/3)y+ 4/9)= -103+ 144+ 4= 45. That is the same as (x- 4)^2+ (y- 2/3)^2= 5. That is a circle with center at (4, 2/3) and radius sqrt(5). Knowing the radius of the circle it is not necessary to do any integration. Its area is 5pi.