If the line x=3 divides the area bounded between the curves y^2=12x and x^2=12y into two parts then the ratio of their areas is

take a look at the graphs

http://www.wolframalpha.com/input/?i=y%5E2%3D12x+and+x%5E2%3D12y

clearly they intersect at (0,0) and (12,12)
height of region = 2√3 x^(1/2) - (1/12)x^2

area from 0 to 3
= ∫ (2√3 x^(1/2) - (1/12)x^2) dx from 0 to 3
= [ 4√3/3 x^(3/2) - (1/36)x^3 ] from 0 to 3
= 4√3/3 (3√3) - 27/36 - 0
= 12 - 3/4 = 45/4

area from 0 to 12
= ∫ (2√3 x^(1/2) - (1/12)x^2) dx from 0 to 12
= [ 4√3/3 x^(3/2) - (1/36)x^3 ] from 0 to 12
= 4√3/3 (12)^(3/2) - (1/36)(1728)
= 96 - 48 = 48

so area from 3 to 12 = 48 - 45/4 = 147/4

ratio of areas = (45/4) : (147/4)
= 45 : 147
= 15 : 49

what do you mean by "their areas"?

Stupid me. I read the question as "two equal parts" and wondered what the fuss was about. Clearly the ratio would be 1:1

Nice catch, Reiny.

Pls also show the diagram

To find the ratio of the areas, we first need to determine the points of intersection between the two curves.

Let's start by equating the two equations:

y^2 = 12x (equation 1)
x^2 = 12y (equation 2)

To eliminate one variable, we can rewrite equation 1 in terms of y and substitute it into equation 2:

(y^2)/12 = 12y

Now, multiply both sides by 12 to get:

y^3 = 144y

Rearranging:

y^3 - 144y = 0

Factoring out y:

y(y^2 - 144) = 0

Setting each factor equal to zero:

y = 0 (which corresponds to the point (0,0))
y^2 - 144 = 0

Solving for y:

y = ±√144 = ±12

So the two curves intersect at three points: (0, 0), (12, 12), and (-12, 12).

Now, let's determine the equation of the area between the curves.

First, let's find the area of the region above the x-axis. The line x = 3 divides this region into two parts.

To find the area of the region between y^2 = 12x and x = 3, we need to find the limits of integration for y. The lower limit is the curve y^2 = 12x, and the upper limit is the line x = 3.

Setting y^2 = 12x equal to x = 3, we get:

12x = 9

x = 9/12 = 3/4

So the limits of integration for x are from x = 0 to x = 3/4.

To find the corresponding limits for y, we solve the equation y^2 = 12x for y:

y = ±√(12x)

Since we are finding the area above the x-axis, we take the positive square root:

y = √(12x)

Now we integrate √(12x) with respect to x from 0 to 3/4:

Integrate[√(12x), {x, 0, 3/4}]

This gives us the area of the region above the x-axis.

To find the area of the region below the x-axis, we consider the same limits of integration but take the negative square root:

y = -√(12x)

Now we integrate -√(12x) with respect to x from 0 to 3/4:

Integrate[-√(12x), {x, 0, 3/4}]

This gives us the area of the region below the x-axis.

Finally, the ratio of the areas is the absolute value of the area below the x-axis divided by the area above the x-axis:

Ratio = |Area below x-axis| / Area above x-axis

So, once you solve the integrals for the respective areas, you can find the ratio by dividing the absolute value of the area below the x-axis by the area above the x-axis.