Answer to Question #101352 in Calculus for Joseph Cielo

Question

Double Integral

1. Integral from 0 to 3, Integral from 0 to 1 of 2x ( square root of x^2 +y dx dy

2. Integral from 0 to ln 3, Integral from 0 to 1 of xye^xy^2 dy dx

Answer 1

$1.\int\limits_0^3(\int\limits_0^12x\sqrt{x^2+y}\,dx)\,dy=\| Let~us~introduce~a~ new~ variable~ \\t=x^2+y.~On ~this~step ~we~ can~think,~ that~ y ~is~ a ~constant, ~because~we~\\integrate~with~respet~to~x\Rightarrow~dt=2xdx\Rightarrow dx=dt/(2x)\|=\int\limits_0^3(\int\limits_{y}^{1+y}\sqrt{t}\,dt)\,dy=\\=\int\limits_0^3\frac{2}{3}((1+y)^{3/2}-y^{3/2})\,dy=\|d(y+1)=dy\|=\frac{2}{3}\int\limits_0^3(1+y)^{3/2}\,d(y+1)-\\ -\frac{2}{3}\int\limits_0^3y^{3/2}\,dy=\frac{2}{3}(\frac{2}{5}(1+y)^{5/2})|^3_{0}-\frac{2}{3}(\frac{2}{5}y^{5/2})|^3_0=$

$=\frac{2}{3}*\frac{2}{5}(4^{5/2}-1^{5/2}-(3^{5/2}-0^{5/2}))=\frac{4}{15}(2^5-1-9\sqrt3)=\frac{4}{15}(31-9\sqrt3)$

$2.\int\limits_0^{\ln3}(\int\limits_0^1xy*e ^{xy^2}\,dy)\,dx=\|(e^{xy^2})'_y=2xy*e^{xy^2}\|=\int\limits_0^{\ln3}1/2 \int\limits_0^12xy*e^{xy^2}\,dy\,dx=\\=\int\limits_0^{\ln3}1/2(e^{xy^2})|_0^1\,dx=1/2\int\limits_0^{\ln3}(e^{x}-1)\,dx=1/2(e^x-x)|_0^{\ln3}=\\=1/2(3-\ln3-(e^0-0))=1/2(2-\ln3)=1-\frac{\ln3}{2}$