Animated Solution for Mathematics - Definite Integration: The area of the region $\{(x,y):x^2\le y\le 8-x^2,y\le 7\}$ is

Visualizing the Region

Imagine you are standing on a coordinate plane, looking at two curves that define a beautiful, enclosed space. We have an upward-opening parabola, $y=x^2$, acting as the floor, and a downward-opening parabola, $y=8-x^2$, acting as the roof.

But there is a twist—a horizontal line at $y=7$ that acts as a ceiling, chopping off the top of our enclosure. This is the region we need to measure.

The area is bounded by $x^2\le y\le 8-x^2$ and $y\le 7$. It is a symmetric, elegant shape, and our goal is to find its total area.

Symmetry and Integration

In JEE Advanced, time is your most precious resource. When you see a region symmetric about the $y$-axis, your first instinct should be to exploit that symmetry.

We will calculate the area in the first quadrant (where $x\ge 0$) and simply multiply the result by 2.

Furthermore, look at the boundaries. If we integrate with respect to $x$, we would need to split the region into multiple parts. However, if we integrate with respect to $y$, the boundaries are much cleaner. We just need to express $x$ in terms of $y$.

The Lower Section ($0\le y\le 4$)

Let's start from the bottom. From $y=0$ to $y=4$, the right boundary of our region is defined by the upward parabola $y=x^2$. Rearranging this, we get $x=\sqrt{y}$.

The area of this lower section, $A_1$, is given by the integral:

Evaluating this is straightforward. The integral of $y^{1/2}$ is $\frac{2}{3}{y}^{3/2}$. Applying the limits from 0 to 4:

This is the area of the lower half in the first quadrant. Keep this value safe; we will need it soon.

The Upper Section ($4\le y\le 7$)

Now, we move to the upper section. From $y=4$ to $y=7$, the right boundary switches to the downward parabola $y=8-x^2$. Rearranging for $x$, we get $x=\sqrt{8-y}$.

The area of this upper section, $A_2$, is:

Here, we must be careful with the chain rule. The integral of $(8-y{)}^{1/2}$ is $-\frac{2}{3}(8-y{)}^{3/2}$. Applying the limits from 4 to 7:

See how the negative signs cancel out beautifully? That is the elegance of calculus at work.

We have the two pieces of our puzzle: $A_1=\frac{16}{3}$ and $A_2=\frac{14}{3}$. The total area in the first quadrant is the sum of these two:

Finally, we apply our symmetry argument. Since the total area $A$ is twice the area in the first quadrant:

The final answer is 20 square units. You have successfully navigated the boundaries, exploited the symmetry, and executed the integration. This is how you master JEE math—not by memorizing formulas, but by understanding the geometry and the logic behind every step.