# Bringing a constant out of double integral (polar coordinates)

Question is: Using polar coordinates, evaluate the integral $\iint\sin(x^2+y^2)dA$ where R is the region $1\le x^2+y^2\le81$

My work for the inside integral, using bounds 1 to 9, was a constant:
$$\int_1^9\sin(r^2)rdr=-\frac{1}{2}\left(\cos81-\cos1\right)$$

Does this mean I can conclude my answer is this constant? In other words, can I bring it out of the integral with respect to theta from 0 to 2$\pi$? I am guessing not because my answer was not correct. If not, could someone explain why?

• Try using MathJax to make your question more readable. – user546997 Apr 25 '18 at 2:44
• Yes you can bring it out. The answer should be correct. What number are you getting for your answer? – John Doe Apr 25 '18 at 2:51
• @JohnDoe The answer is that constant (my answer) times 2pi. – h.jb Apr 25 '18 at 2:55
• That's the same answer I get. What is the "correct" answer. – saulspatz Apr 25 '18 at 3:01
• I get $-0.74...$. What number are you getting? Do you have an actual number? That would help us figure out what your issue is – John Doe Apr 25 '18 at 3:03

You ignored the angular part of the integral: $$\int\int_R\sin(x^2+y^2)\,dx\,dy=\int_0^{2\pi}\int_1^9r\sin(r^2)\,dr\,d\theta\\=\left(\int_0^{2\pi}\,d\theta\right)\left(\int_1^9r\sin(r^2)\,dr\right)=2\pi\times(-0.118\dots)=-0.74\dots$$
• @h.jb yes, we are $$\int_0^{2\pi}\,d\theta=\int_0^{2\pi}1\,d\theta=[\theta]_0^{2\pi}=2\pi$$Does that make sense now? We can split the integral in this way because there are no terms with $\theta$ in the original integral. – John Doe Apr 25 '18 at 3:22