Evaluate ∬(x^2+y^2)^2/(x^2y^2) dx dy over the region common to the circles x^2+y^2=7x and x^2+y^2=11y.

Answer 1

Let
C1: x^2+y^2=7x <=> (x+3.5)^2+y^2=3.5^2
C2: x^2+y^2=11y <=> x^2+(y+5.5)^2=5.5^2

We see that C1 has a centre at (-3.5,0), and C2 has a centre at (0,-5.5)

Thus (0,0) is one of the intersection points.

To find the other intersection point, we equate C1=C2 to get 7x=11y (common chord).

Substitute y=7x/11 in C1 or C2 to get the other intersection point of x=-847/170, and y=-539/170.
Thus (-847/170,-539/170) is the other intersection point.

Now we would express the parts of the arcs which intersect, Y1(x) and Y2(x).

Solving for y in C1 below the x-axis,
Y1(x)=-√(3.5^2-(x+3.5)^2)

and solving for y in C2 to the left of the y-axis,
Y2(x)=-√-5.5+√(5.5^2-x^2)

Thus the required area is obtained by integrating Y2(x)-Y1(x) between x=-847/170 and 0.

I get Area = 10.19.

Answer 2

Typo:
Y2(x)=-5.5+√(5.5^2-x^2)

Subsequent lines are not affected.

Answer 3

But we are taking a double integral. This would be finding a volume.

Answer 4

Indeed.
I have misread the question and just calculated the area.

Some work will need to be done on the function F(x,y)=(x^2+y^2)^2/(x^2*y^2) because it is an improper integral as (x,y)->(0,0). The existence of the integral will have to be demonstrated at (0,0), and the improper integral evaluated accordingly. Note the fact that even if F(x,y) does not exist at (0,0), the integral may exist.

Please kindly repost the question to get the quickest possible response. Sorry!
If I have any new information, I will answer the new post.

Answer 5

ok, thanks!