Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region bounded by the curves $x^2+2y-1=0$, $y^2+4x-4=0$ and $y^2-4x-4=0$, in the upper half plane is ____.

A Triple-Parabola Dance

Imagine you are standing on a coordinate plane, looking at three distinct curves. We have $C_1:x^2+2y-1=0$, $C_2:y^2+4x-4=0$, and $C_3:y^2-4x-4=0$.

$C_1$ is a classic downward-opening parabola, resting its vertex at $(0,0.5)$ and kissing the x-axis at $x=\pm 1$.

$C_2$ and $C_3$ are side-opening parabolas. $C_2$ opens to the left with its vertex at $(1,0)$, and $C_3$ opens to the right with its vertex at $(-1,0)$. They meet at $(0,2)$. This is a geometric puzzle where these three curves carve out a beautiful, symmetric region in the upper half plane.

When you see a problem like this, your first instinct might be to dive straight into the integration. But wait! Look at the symmetry.

The entire region is perfectly mirrored across the y-axis. This is a gift.

Instead of calculating the area of the entire region, we can focus our energy on the first quadrant (where $x\ge 0$) and simply double the result. This simple realization cuts our workload in half and reduces the chance of a sign error.

The Subtraction Method

Now, how do we calculate the area in the first quadrant? The region is bounded by the y-axis, $C_1$, and $C_2$.

If we try to integrate with respect to $x$ for the whole thing, we would have to split the integral because the upper boundary changes. Instead, let's use a more elegant approach: the subtraction method.

We calculate the area under $C_2$ (with respect to $y$) from $y=0$ to $y=2$. This gives us the entire area between $C_2$ and the y-axis. Then, we subtract the area under $C_1$ (with respect to $x$) from $x=0$ to $x=1$. This removes the 'unwanted' space below $C_1$.

Execution

Let's set up our integral. The area in the first quadrant is given by:

Substituting our expressions, we get:

Let's compute these one by one. For the first part:

For the second part:

Finally, we subtract these values:

Since we are looking for the total area, we multiply by 2 to account for the symmetry:

The area is exactly 2 square units. It is a clean, satisfying result that rewards our careful planning. Remember, in JEE Advanced, the most complex problems often yield to the simplest, most elegant strategies.