calculate $\int_Df(x,y)dxdy$ in these cases

Question

• $f(x,y)={1\over{x^2y}}$ With $D=${$ x\ge1,{1\over{x}}\le y\le x $ }
• $f(x,y) = e^{2x+y}$ With $D=${ $x\le a ,x+y\le a $ }

Answer 1

Write the double-integrals as iterated integrals: $$ \iint_D \frac{1}{x^2y}\,\mathrm dy\,\mathrm dx=\int_1^\infty\int_{1/x}^x \frac{1}{x^2y}\,\mathrm dy\,\mathrm dx $$ and $$ \iint_D e^{2x+y}\,\mathrm dy\,\mathrm dx=\int_{-\infty}^a\int_{-\infty}^{a-x} e^{2x+y}\,\mathrm dy\,\mathrm dx. $$ Move on by treating the inner most integral first for a fixed value of $x$. Then calculate the outer most integral.

Answer 2

for the first one you have $\int_1^\infty \int_{\frac{1}{x}}^x \frac{1}{x^2y} dydx=\int_1^\infty\frac{1}{x^2}(ln|x|-ln|\frac{1}{x}|)dx=\int_1^\infty \frac{1}{x^2} ln(x^2)=-\frac{ln(x^2)+2}{x}|_1^\infty$. At $\infty$ the expression equals 0 which you can get by applying L'Hopital's rule twice. At 1, the expression equals -2, which gives you 2 as a result. for the second one you have $\int_{-\infty}^a\int_{-\infty}^{a-x} e^{2x}e^y dydx=\int_{-\infty}^ae^{2x}\int_{-\infty}^{a-x}e^ydydx=\int_{-\infty}^a e^{2x}e^{a-x}dx=\int_{-\infty}^a e^{a+x}dx=e^a\int_{-\infty}^ae^xdx=e^a \cdot e^a=e^{2a}$