Area between two curves in the first quadrant

Key concept: This is a neat first-quadrant area problem because the boundary is easier to read with horizontal slices. Once the intersection point is found, the integral becomes very manageable.

Step by step

1) Find the intersection

The curves are

$x=y\sqrt{y^4+1}$

and

$x=2y.$

Set them equal:

$y\sqrt{y^4+1}=2y.$

In the first quadrant, $y\ge 0$, so either $y=0$ or

$\sqrt{y^4+1}=2 \Rightarrow y^4=3 \Rightarrow y=3^{1/4}.$

So the region runs from $y=0$ to $y=3^{1/4}$.

2) Write the area integral

Right curve minus left curve:

$A=\int_0^{3^{1/4}}\left(2y-y\sqrt{y^4+1}\right)dy$

3) Evaluate

For the first term:

$\int_0^{3^{1/4}}2y\,dy = \left[y^2\right]_0^{3^{1/4}}=\sqrt{3}$

For the second term, use $t=y^2$, so $dt=2y\,dy$:

$\int y\sqrt{y^4+1}\,dy = \frac12\int \sqrt{t^2+1}\,dt$

with limits $t=0$ to $t=\sqrt3$.

That gives

$\frac12\cdot \frac14\left(t\sqrt{t^2+1}+\ln\left(t+\sqrt{t^2+1}\right)\right)_0^{\sqrt3}$

$=\frac18\left(2\sqrt3+\ln(2+\sqrt3)\right).$

So the area is

$A=\sqrt3-\frac18\left(2\sqrt3+\ln(2+\sqrt3)\right)$

$=\frac{3\sqrt3}{4}-\frac18\ln(2+\sqrt3).$

Pitfall alert

A common slip is to integrate with respect to x even though both curves are already written as x-functions of y. Using horizontal slices is much cleaner here. Another mistake is missing the first-quadrant restriction and forgetting that y starts at 0.

Try different conditions

If the right boundary were changed from $x=2y$ to $x=ky$, the intersection would satisfy $\sqrt{y^4+1}=k$, so the same method still works. Only the upper limit changes, and the rest of the setup is identical.

Further reading

definite integral, intersection point, area between curves