Find the area bounded by curves

[gruba]

Homework Statement

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

Homework Equations

-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]

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.