http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Thu, 17 Jul 2014 11:12:06 -0500http://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?Thu, 17 Jul 2014 11:07:00 -0500http://calc3.askbot.com/question/120/double-integrals-from-158/http://calc3.askbot.com/question/120/double-integrals-from-158/?answer=121#post-id-121Okay, I just realized that I didn't change by bounds of integration when I did my $u$ substitution. It should be: $$1/3 \int_1^2 \sqrt{u} \hspace{2 mm} du$$ $$1/3(2/3 \sqrt{u^3} ) \biggr|_1^2$$ $$=4/9 \sqrt{2}-2/9$$ simple mistake!Thu, 17 Jul 2014 11:12:06 -0500http://calc3.askbot.com/question/120/double-integrals-from-158/?answer=121#post-id-121