Thread: definite integral of square root of 1-x^2 +2 using geometric consideration

1. definite integral of square root of 1-x^2 +2 using geometric consideration

Question states, using geometric considerations, what does $\displaystyle \int\limits_0^1 {\sqrt {1 - x^2 } } + 2dx$ equal

I can see that between zero and one that sqrt (1-x^2) is quadrant of a circle so the area would be 1/4*pi*r^2 but where does the +2 come in, I see it would move the circle up the y axis by 2. But I'm stuck

2. Originally Posted by Craka
Question states, using geometric considerations, what does $\displaystyle \int\limits_0^1 {\sqrt {1 - x^2 } } + 2dx$ equal

I can see that between zero and one that sqrt (1-x^2) is quadrant of a circle so the area would be 1/4*pi*r^2 but where does the +2 come in, I see it would move the circle up the y axis by 2. But I'm stuck
If $\displaystyle y= \sqrt{1- x^2}+ 2$ then $\displaystyle y- 2= \sqrt{1- x^2}$ so $\displaystyle (y-2)^2= 1- x^2$ and, finally, $\displaystyle x^2+ (y-2)^2= 1$. that is a circle with center at (0, 2) and radius 1. Since the square root is positive, you have the upper half of that circle and since x is from 0 to 1, rather than -1 to 1, you have only the upper right quadrant of that circle. The integral is, of course, $\displaystyle \pi/4$, the area of 1/4 of a circle of radius 1.

3. wow some things look so simple when laid out. Thanks