Finding the area under a curve with respect of both $y$ and $x$

Question

I have the following set of equations

$$y = 1$$ $$y = x ^{1/4}$$

from $x = 0$, to $x = 1$

Now, I understand that the easiest way to do this is using the $x$ (question: because we are cutting vertically right?)

$$\int_0^1{1-(x^{1/4})}$$

However, is it possible to find the area under the curve with respect to $y$?

Answer 1

Of course it is possible: $$ \int_0^1 y^4 dy =\frac{1}{5}. $$ The other integral is $$ \int_0^1 x^{1/4} dx =\frac{4}{5}. $$

The equality $\int_0^1 y^4 dy+\int_0^1 x^{1/4} dx=1$ is of course not just a coincidence.