Answer to Question #348505 in Calculus for Migz

Question

Two curves having an equation of x=2√y and y=2√x intersect each

other. Compute the area between the two curves.

Answer 1

Let's rewrite the function $x=2\sqrt{y}.$

$x^2=(2\sqrt{y})^2, x\geq 0, y\geq 0;$

$x^2=4y, x\geq 0, y\geq 0;$

$y=\frac{1}{4}x^2, x\geq 0, y\geq 0.$

Let's find the interseptions of the curves:

$\frac{1}{4}x^2=2\sqrt{x},$

$(\frac{1}{4}x^2)^2=(2\sqrt{x})^2,$

$\frac{1}{16}x^4=4x,$

$x^4=64x,$

$x^4-64x=0,$

$x(x^3-64)=0,$

$x=0$ or $x^3-64=0,$

$x=0$ or $x=4.$

$A=\int_0^42\sqrt{x}dx-\int_0^4 \frac{1}{4}x^2dx=$

$=2\int_0^4\sqrt{x}dx- \frac{1}{4}\int_0^4x^2dx=$

$=2\cdot \frac{2}{3}x\sqrt{x} |_0^4-\frac{1}{4}\frac{x^3}{3}|_0^4=$

$=\frac{4}{3}\cdot 4\sqrt{4}-\frac{1}{4}\cdot \frac{4^3}{3}|_0^4=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}=5\frac{1}{3}.$

Answer: $5\frac{1}{3}.$