http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Mon, 21 Jul 2014 10:17:19 -0500http://calc3.askbot.com/question/147/improper-double-integrals/In section 15.2, question #19a, it asks: Consider the integral $\int\int 1/\sqrt{x^2+y^2} dA$, where $D$ is the unit disk centered at the origin. a) Why might this integral be considered improper? The link to see this graph did not work for me, but I used wolfram alpha and the graph looked kind of like a piece of paper being folded upwards. I don't know why this could be considered improper? Any thoughts?AnonymousMon, 21 Jul 2014 10:17:19 -0500http://calc3.askbot.com/question/147/http://calc3.askbot.com/question/125/setting-up-double-integrals/I am working on #16 from 15.8 and I'm wondering if I have set up my integrals correctly. The question asks: Evaluate $\int\int x^2 \hspace{2 mm} dA$ over the region in the first quadrant bounded by the hyperbola $xy=16$ and the lines $y=x, y=0$, and $x=8$. I came up with $$\int_4^8\int_{16/x}^x x^2\hspace{2 mm} dydx$$ is this correct?AnonymousThu, 17 Jul 2014 12:15:19 -0500http://calc3.askbot.com/question/125/http://calc3.askbot.com/question/120/double-integrals-from-158/So I am having trouble with #9 from 15.8: Compute $$\int_0^1\int_\sqrt{y}^1 \sqrt{x^3+1} \hspace{2 mm} dxdy$$ here is my work $$\int_0^1\int_0^{x^2} \sqrt{x^3+1} \hspace{2 mm} dydx$$ $$\int_0^1 y\sqrt{x^3+1} \biggr|_0^{x^2} dx$$ $$\int_0^1 x^2\sqrt{x^3+1} \hspace{2 mm} dx$$ I used $u$ substitution here $\hspace{14 pc}$ Let $ u=x^3+1$ $$du=3x^2 dx$$ $$1/3 du=x^2 dx$$ $$1/3 \int_0^1 \sqrt{u} \hspace{2 mm} du$$ $$1/3(2/3 \sqrt{u^3}) \biggr|_0^1$$ $$=2/9$$ what am I missing here?AnonymousThu, 17 Jul 2014 11:07:00 -0500http://calc3.askbot.com/question/120/