Bounded Area between $y=x^2$, and $y=9$, and $y=k$

Question

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?

Answer 1

Yes, correct as you had found. $$\int_\limits{0}^9\sqrt{y} dy = 18$$ $$\dfrac{18}{2}=\int_\limits{k}^9\sqrt{y}dy \rightarrow k=.. $$

Answer 2

1 year and 8 months later Update

I ended solving the problem setting the integral in terms of $dx$ which gave a another possible method to engage the problem. It definitely was something that I didn't think about at that heat of the moment. Here it is.

X-axis Method

\begin{align}I&=\int_{0}^{3}(9-x^2)dx=18 \end{align} Since the k is slicing the area into equal parts then I can say that: \begin{align}\int_{0}^{\sqrt{k}}(k-x^2)dx&=9 \\ [k^{\frac{3}{2}}-\frac{k^{\frac{3}{2}}}{3}]&=9 \\ \frac{2}{3}k^{\frac{3}{2}}&=9 \\ k=(\frac{27}{2})^{\frac{2}{3}} \end{align}