1. ## sketch regions

Sketch the region bounded by the semicircle y=sqrt(2x-x^2)
and the x-axis and then evaluate the given integral.........

then the class notes say therefore this is "a circle centred at (1,0) with radius 1.

how do you know this is a circle with radius one?
how do you know where it is centred?
starting a table of values is the only way I know to work out how to plot this

2. Originally Posted by samdmansam
Sketch the region bounded by the semicircle y=sqrt(2x-x^2)
and the x-axis and then evaluate the given integral.........

then the class notes say therefore this is "a circle centred at (1,0) with radius 1.

how do you know this is a circle with radius one?
how do you know where it is centred?
starting a table of values is the only way I know to work out how to plot this
You could do this:

$y=\sqrt{2x-x^2}\implies y^2=2x-x^2$

Then we see that $x^2-2x+y^2=0$

Completing the square gives us $x^2-2x+1+y^2=1\implies (x-1)^2+y^2=1$

Now comparing this to the standard form of a circle: $(x-x_0)^2+(y-y_0)^2=r^2$, where $(x_0,y_0)$ is the center of the circle and $r$ is the radius of the circle, we now see that $(x-1)^2+y^2=1$ is centered at $(1,0)$ and has a radius of one.

Now note that we are not graphing the entire circle. Its just the upper half, since that is hinted by the positive square root. If it was $-\sqrt{2x-x^2}$, that would be dealing with the semicircle below the x axis.

Now, if we were to integrate this guy, it would require some extra thinking:

$\int_0^2\sqrt{2x-x^2}\,dx$

This is the same thing as saying $\int_0^2\sqrt{1-(x-1)^2}\,dx$

Now apply the trig substitution $x-1=\sin\vartheta$

Can you take it from here?

--Chris