Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$ and $y = 4$ into two regions with equal area.
I could find the area of this region but i have no clue how to split it parallel to the $x$ axis perfectly… this is too abstract for me…
I found the whole area and got $\dfrac{32}{3}$… I tried plugging that back into the definite integral but that definitely wasn't right…
Best Answer
We can find the area by integrating with respect to $x$ or with respect to $y$.
With respect to $y$ is easier. I would prefer to use symmetry and say that the area is equal to $$2\int_0^4 y^{1/2}\,dy.$$ Integrate. We get $(2)(2/3)4^{3/2}=32/3$, your answer.
So we want the area from $y=0$ to $b$ to be half of $\frac{32}{3}$, that is, $\frac{16}{3}$.
The area from $y=0$ to $y=b$ is equal to $$2\int_0^b y^{1/2}\,dy.$$ This is equal to $(2)(2/3)b^{3/2}$.
Thus we need to find the $b$ such that $$(2)(2/3)b^{3/2}=\frac{16}{3}.$$ Simplify. We get the equation $b^{3/2}=4$.
Solve for $b$. We get $b=4^{2/3}$. This can be written in various other ways, such as $b=2\sqrt[3]{2}$.
Remark: We could have bypassed the initial computation of the area, and written that we want to find $b$ such that $$2\int_0^b y^{1/2}\,dy=2\int_b^4 y^{1/2}\,dy.$$ But I think it is better to find first the area from $0$ to $4$, to get some concrete feeling about the situation.
For a "reality check" note that our answer must be $\gt 2$: we have to go more than half of the way to $4$ to capture half the area.