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/135/finding-volume-using-integration/I have been thinking about this problem in section 15.1 #30: Three cylinders of radius 1 intersect at right angles at the origin, as shown in figure 15.1.6. Find the volume contained inside all three cylinders. My first thought is to find the points of intersection but other than that I'm pretty much lost. Any ideas?AnonymousFri, 18 Jul 2014 09:18:08 -0500http://calc3.askbot.com/question/135/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/http://calc3.askbot.com/question/118/double-integrals/So with the problems he put on the board at the end of class today, I worked out the first double integral and was just hoping someone could verify if I'm doing it right. So he gave us the domain :  and the double integral as: $$\int\int x^2y\delta A$$ for which I came up with the integrals being: $$\int _0 ^2 \int _0 ^{4-x^2} x^2y \delta y \delta x$$ $$=\int _0 ^2 \frac 12x^2y^2\delta x \mid _0 ^{4-x^2}$$ $$=\int _0 ^2 \frac 12 x^2 (4-x^2)^2 \delta x$$ $$= \int _0 ^2 \frac 12 x^2(16-8x^2+x^4)\delta x$$ $$=\int _0 ^2 8x^2 -4x^4 + \frac 12 x^6 \delta x$$ $$=\frac 83 x^3 - \frac 45 x^5 + \frac {1}{14} x^7 \mid _0 ^2$$ $$= \frac 83 (8) - \frac 45(32) + \frac{1}{14}(128)$$ $$= \frac{64}{3} -\frac{128}{5} +\frac{128}{14}$$ This method seems to make sense to me(hoping that my terms of integration are correct), so I'm just hoping that someone else follows this too, or you could explain and easier way to do it. Thanks!TiffanyThu, 17 Jul 2014 10:21:11 -0500http://calc3.askbot.com/question/118/