Animated Solution for Mathematics - Definite Integration: The area of the region $S=\{(x,y):y^2\le 8x,y\ge \sqrt{2}x,x\ge 1\}$ is

Imagine you are standing on the Cartesian plane, looking at a landscape defined by three simple rules. We have a parabola, $y^2=8x$, which is a classic curve that opens wide to the right, growing steadily as $x$ increases.

Then, we have a straight line, $y=\sqrt{2}x$, which slices through the origin like a sharp blade. Finally, we have a vertical wall at $x=1$.

The region $S$ is the space trapped between these three entities. To find the area of this region, we are essentially calculating the accumulation of infinite, tiny vertical strips, each with a height equal to the difference between the 'roof' and the 'floor' of our region.

This is the heart of integral calculus—turning a complex shape into a sum of simple rectangles.

The Moment of Truth

Before we can integrate, we must know where our region begins and ends. We know it starts at $x=1$ because of the constraint $x\ge 1$.

But where does it end? The parabola and the line meet at specific points. To find them, we set the equations equal to each other.

Substituting $y=\sqrt{2}x$ into $y^2=8x$, we get $(\sqrt{2}x{)}^2=8x$, which simplifies to $2x^2=8x$. Rearranging this gives $2x^2-8x=0$, or $2x(x-4)=0$.

This tells us they intersect at $x=0$ and $x=4$. Since our region is constrained to $x\ge 1$, the upper boundary of our region is clearly $x=4$. We have our limits: from $x=1$ to $x=4$.

Now, let's build our integral. The area $A$ is the integral of the upper curve minus the lower curve.

The upper curve is the parabola, which we write as $y=2\sqrt{2}\sqrt{x}$ (taking the positive root for the upper half), and the lower curve is the line $y=\sqrt{2}x$. So, our integral becomes:

Notice how we can factor out the constant $\sqrt{2}$ to make the algebra cleaner:

This is the moment where the physics of the problem meets the elegance of pure mathematics.

Now, we apply the power rule for integration. For the first term, $2x^{1/2}$, the integral is $2\cdot \frac{x^{3/2}}{3/2}=\frac{4}{3}{x}^{3/2}$. For the second term, $x$, the integral is $\frac{x^2}{2}$.

Putting it all together, we have:

First, we evaluate at the upper limit $x=4$:

Next, we evaluate at the lower limit $x=1$:

Finally, we subtract the lower limit value from the upper limit value:

And there it is! The area of our region is $\frac{11\sqrt{2}}{6}$ square units.

It is a beautiful result, isn't it? The key to this problem wasn't just the integration, but the visualization—understanding the boundaries and respecting the constraints.

Whenever you face a problem like this, remember to sketch it out, identify your 'roof' and 'floor', and let the calculus do the heavy lifting. You have the tools; now go out and conquer the next one!