Find the horizontal line $y=k$ which divides the enclosed area between the curves $y=x^2$ and $y=16$ in two equal parts.
I have sketched the graph and concluded mistakenly that: $$\int_{-4}^4 16-k\, dx + \int_{-4}^4 k-x^2\, dx = \int_{-4}^4 x^2\, dx$$
Any help is appreciated.
Best Answer
Try to consider the following integrals instead: $$\int_0^k (+\sqrt{y}-(-\sqrt{y}))dy=\int_k^{16}(+\sqrt{y}-(-\sqrt{y}))dy$$