Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am working on the following integral problem:

$$\int_{0}^{1}(\sqrt{2-x^2}-\sqrt{2x-x^2})dx$$

There is a hint as well, which suggests interpreting the definite integral as the area bounded by appropriate curves. I have graphed both the curves with Wolfram Alpha (link), and I understand the area I am trying to find. I am stuck on solving the integral. The tools I have been provided with so far are fairly basic such as substitution and integration by parts.

share|improve this question
6  
For first integral, let $x=\sqrt{2}\sin\theta$. For second, note that $2x-x^2=1-(1-x)^2$ and let $1-x=\sin\phi$. –  André Nicolas Jan 12 '12 at 5:17

1 Answer 1

up vote 3 down vote accepted

André has given you a calculus solution; here’s one that uses no calculus.

The graph of $y=\sqrt{2-x^2}$ is the upper half of a circle of radius $\sqrt2$ centred at the origin; that of $y=\sqrt{2x-x^2}=\sqrt{1-(x-1)^2}$ is the upper half of a circle of radius $1$ centred at $(1,0)$. Thus, $$\int_0^1 \sqrt{2x-x^2} dx=\frac{\pi}4\;,$$ a quarter of the area of a circle of radius $1$. We can also get $$\int_0^1 \sqrt{2-x^2} dx$$ without calculus, but it requires a little more cleverness. If you sketch the quarter-disk in the first quadrant bounded by $y=\sqrt{2-x^2}$, you’ll see that it’s the union of the unit square $$S=\{(x,y):0\le x,y\le 1\}\;,$$ the region $T$ bounded by the $y$-axis, the line $y=1$, and the curve $y=\sqrt{2-x^2}$, and the region $R$ bounded by the $x$-axis, the line $x=1$, and the curve $y=\sqrt{2-x^2}$. Regions $T$ and $R$ are clearly congruent, so they have the same area. The area of the whole quarter-disk is $\pi/2$, so $$\frac{\pi}2=1+2\operatorname{area}(T)\;,$$ and $$\operatorname{area}(T)=\frac12\left(\frac{\pi}2-1\right)=\frac{\pi-2}4\;.$$ Finally, $$\int_0^1 \sqrt{2-x^2} dx=\operatorname{area}(S)+\operatorname{area}(T)=1+\frac{\pi-2}4=\frac{\pi+2}4\;,$$ and $$\int_{0}^{1}\left(\sqrt{2-x^2}-\sqrt{2x-x^2}\right)dx=\frac{\pi+2}4-\frac{\pi}4=\frac12\;.$$

share|improve this answer
2  
@J.Borges: If you are interested in the history of mathematics, you might like to know that this is essentially the famous quadrature of a lune by Hippocrates. –  André Nicolas Jan 12 '12 at 5:52
    
Looking at the figure described in the first three lines one immediately sees that the area in question is $\bigl(({1\over 8}2\pi+{1\over 2}\bigr) - {\pi\over 4}={1\over 2}$. –  Christian Blatter Jan 12 '12 at 14:14
    
Thank you very much for your help Brian, and thank you very much André for the link (I'll have to wait until Wikipedia is back online, however). –  J.Borges Jan 18 '12 at 19:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.