Use a double integral to compute the area of region R

Question

R is bounded by the graphs of $ y = x^3, x = 0, y = 1$

I sketched the graph of $x^3$ and $y = 1$ but I am not sure how to build the integral from here.

This is what I got for the bounds $$ \int_0^1\ \int_0^{x^3} [f(x,y) + g(x,y)] dA $$ since there are three curves to binding $\mathbb{R}$ which function is $f$ and which $g$?

Answer 1

Your plot looks like this:

Your curve is $y=x^3$ or similarly $x = y^{1/3}$.

If you want to integrate in $x$ first, then $0 \le x \le y^{1/3}$ with $y \in [0,1]$, while if you integrate $y$ first, then $x^3 \le y \le 1$ while $x \in [0,1]$. So you end up with $$ A = \int_0^1 \int_0^{y^{1/3}} dxdy = \int_0^1 \int_{x^3}^1 dydx. $$

UPDATE

To evaluate, the first one yields $$ A = \int_0^1 \int_0^{y^{1/3}} dxdy = \int_0^1 y^{1/3} dy = \left. \frac{y^{4/3}}{4/3} \right|_0^1 = \frac{3}{4}. $$

The second one yields $$ A = \int_0^1 \int_{x^3}^1 dydx = \int_0^1\left[1 - x^3\right] dx = \left[x - \frac{x^4}{4}\right]_0^1 = 1 - \frac{1}{4} = \frac{3}{4}. $$

Answer 2

What you need to compute is$$\int_0^1\int_{x^3}^11\,\mathrm dy\,\mathrm dx,$$since $x$ can take any value between $0$ and $1$ and, for each $x$, $y$ can take any value between $x^3$ and $1$.