# Find the area bounded by curves

## Homework Statement

Find area bounded by functions $y_1=\sqrt{4x-x^2}$ and $y_2=x\sqrt{4x-x^2}$.

-Integration
-Area

## The Attempt at a Solution

From $y_1=y_2\Rightarrow x=1$. Intersection points of $y_1$ and [/itex]y_2[/itex] are $A(0,0),B(1,\sqrt 3),C(4,0)$. Domain of $y_1$ and $y_2$ is $x\in [0,4]$. On the interval $x\in[0,1]\Rightarrow y_1\ge y_2$ and on the interval $x\in[1,4]\Rightarrow y_1\le y_2$.

$A=\int_0^1 (y_1-y_2)\mathrm dx+\int_1^4 (y_2-y_1)\mathrm dx=\int_0^1 (1-x)\sqrt{4x-x^2}\mathrm dx+\int_1^4 (x-1)\sqrt{4x-x^2}\mathrm dx$

How to solve integrals $\int \sqrt{4x-x^2}\mathrm dx$ and $\int x\sqrt{4x-x^2}\mathrm dx$?

Substitution $u=\sqrt{\frac{x}{4-x}}\Rightarrow du=\frac{2}{(x-4)^2\sqrt{\frac{x}{4-x}}}dx$ doesn't seems to work.

BvU
Homework Helper
Perhaps u = x-2 ?

By the way, "integration" doesn't qualify as a relevant equation.

Having said that (PF guidelines,...), I need to cheat to find this one. CRC handbook of chem and phys helped.

HallsofIvy
A fairly standard way to do such integrals is to complete the square inside the square root. $4x- x^2= -(x^2- 4x)= -(x^2- 4x+ 4- 4)= -(x- 2)^2+ 4$. So $\int \sqrt{4x- x^2}dx= \int \sqrt{4- (x- 2)^2}dx$ and now make the substitution u= x- 2 so this becomes $\int \sqrt{4- u^2}du$. Wth u= x- 2, x= u+ 2 so the second integral becomes $\int x\sqrt{4x- x^2}dx= \int (u+ 2)\sqrt{4- u^2}du$.