I have been given a function to find the horizontal line $y=k$ which would splice the area bounded between $y=x^2$, and $y=9$, into two equal parts. I approached the function using symmetry to find which $k$ value would splice the bounded region into two equal parts. I used this approach:
Approach
$$\int_\limits{0}^9\sqrt{y} dy = 18$$
Which is half of the given area, however, since the given function is symmetrical I then stated that this area, which I labeled as $A_1$.
$$A_1=\int_\limits{k}^9\sqrt{y}dy +\int_\limits{0}^k\sqrt{y}dy$$
From which I then deduced that half of $A_1$ is the equivalent of one of them from which I choose to do.
$$9=\int_\limits{0}^k\sqrt{y}dy$$
$$9=\frac{2}{3}k^{\frac{3}2}$$
Therefore, I got $k$ leaving it in irrational form:
$$k=(\frac{3}{2}\cdot9)^{\frac{2}3}$$
Is my answer correct?
Best Answer
Yes, correct as you had found. $$\int_\limits{0}^9\sqrt{y} dy = 18$$ $$\dfrac{18}{2}=\int_\limits{k}^9\sqrt{y}dy \rightarrow k=.. $$