# What am I doing wrong in the evaluation of this integral?

So Here is this question : "evaluate $$\iint\sqrt{1-x^2-y^2} \, dx \, dy$$ where the region is bounded by the circle $$x^2+y^2=x$$."

Now if I turn it polar form and take limits of $$r$$ from $$0$$ to $$\cos\theta$$ and limits of $$\theta$$ from $$-\pi/2$$ to $$\pi/2$$, I get to point where I get integrand of $$\sin^3\theta$$, $$\theta$$ varying from $$-\pi/2$$ to $$\pi/2$$, which should be zero, right? As it is an odd function.

But in the book they have taken the limit of $$\theta$$ from $$0$$ to $$\pi$$ and came up with the answer $$\frac{\pi}{3}-\frac{4}{9}$$ which I only get if I put the integrand of $$\sin^3 \theta$$ from $$-\pi/2$$ to $$\pi/2$$ equals twice the integrand of $$\sin^3 \theta$$ from $$0$$ to $$\pi/2$$ which shouldn't be the case because it's an odd function. What am I doing wrong?

• Hello! Welcome to Math.SE. That huge single paragraph is ugly as sin. Please try to format your question more clearly. Here's a quick tutorial on how to write math on this site: meta.math.stackexchange.com/questions/5020/… – Newb Jan 26 '14 at 7:28
• i could learn that later...i need help right now. – TimeLord Jan 26 '14 at 7:30
• I'm trying to help you --- nobody is going to answer your question. I don't even want to try to read that huge ugly block of text. At least split it into some paragraphs. – Newb Jan 26 '14 at 7:36
• okay i am trying,i didn't notice it'd be that difficult...thank you so much xx – TimeLord Jan 26 '14 at 7:38
• I have fixed it. I understand you are new here, but it is considered very poor form to hastily write your question and then expect others to go out of their way to not only reformat the question nicely and make sense of it, but to then answer the question. – heropup Jan 26 '14 at 7:40

The answer in the book is correct: the value of the integral is $\frac{\pi}{3} - \frac{4}{9}$. The appropriate transformation is $$x = r \cos \theta, \quad y = r \sin \theta,$$ which transforms the region of integration to $$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}, \quad 0 \le r \le \cos \theta.$$ This gives the resulting integral $$\int_{\theta = -\pi/2}^{\pi/2} \int_{r = 0}^{\cos \theta} \sqrt{1 - r^2} \, r \, dr \, d\theta = \frac{1}{3} \int_{\theta = -\pi/2}^{\pi/2} 1 - (1-\cos^2\theta)^{3/2} \, d\theta,$$ but your mistake is in writing this latter integrand as $1 - \sin^3 \theta$ rather than $1 - |\sin^3 \theta|$, because $1 - \cos^2 \theta \ge 0$, and without the absolute value, $\sin^3 \theta$ is negative on the subset of the interval $-\frac{\pi}{2} \le \theta < 0$. Thus you cannot use odd/even function reasoning. Instead, you need to use symmetry and evaluate half the integral.