Find the horizontal line which divides the enclosed area between curves

Question

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.

Answer 1

Try to consider the following integrals instead: $$\int_0^k (+\sqrt{y}-(-\sqrt{y}))dy=\int_k^{16}(+\sqrt{y}-(-\sqrt{y}))dy$$

Answer 2

Note: The line $y=k$ does not cut the parabola in -4 or 4. Compute $ \int_{-k}^{k} (k-x^2) dx$. This has to be equal to one half of the given area $\int_{-4}^{4} (16-x^2) dx$.

Can you finish it?

Answer 3

As the graph is symmetrical about the $y$ axis you can ignore the negatives and just work from 0. You have correctly realised that $x$ varies from $0$ to $4$. If you integrate across this area you get the total area under the curve $=\frac{64}{3}$.

The total area in the rectangle is $4 \times 16 = 64$ so the area between the curve and $y = 16$ is

$64-\frac{64}{3} = \frac{128}{3}.$

Since you want to split this in half you have the two regions $A$ and $B$. Where $A$ is the lower section and $B$ is the higher section. You also have the areas are equal: $A = B = \frac{64}{3}.$

You can then calculate $A$ by doing $kx_1 - \int^{x_1}_0 x^2 dx$ where $k = x_1^2$.

Working this through and plugging in your value of $A$ should give you the answer.