Average height in calculus

Question

Calculate the average height above the x-axis of a point in the region $0\leq x\leq2$ and $0\leq y\leq x^2$. But I have no idea what average height is. I have googled it and find the formula $\bar H=\frac1{\text{Area}(D)}\int\int\limits_DydA$. Could anyone tell me why we have a $y$ in the given formula? How does it come from?

Answer 1

For every point in a 2-dimensional figure, $y$ is the height of that point from the x-axis. Thus, when you want to add up heights of every point, you get $\int\int_D ydA$. You also want to divide by $Area(D)$, since you want the average height (and not the sum of heights).

Thus, the answer would be as follows: first, $Area(D) = \int_{0}^{2} \int_{0}^{x^2} dydx = \int_{0}^{2} x^2dx = \frac{8}{3}$. Then, we get that $\int_{0}^{2} \int_{0}^{x^2} ydydx = \int_{0}^{2}\frac{x^4}{2}dx = \frac{16}{5}$. Thus, the average height is $\frac{16*3}{5*8} = \frac{6}{5}$. Cross-check that this makes sense intuitively (i.e, that you would expect it to be more than $1$).

Answer 2

$$\bar{H} = \frac{\int\limits_{x=0}^2 \int\limits_{y=0}^{x^2}\ y\ dy\ dx}{\int\limits_{x=0}^2 x^2\ dx} = \frac{6}{5}$$