Question

okay so I am having issues with double integrals. I need to find the volume bounded below by the rectangle R 1<_x<_2 , 1<_y<_e in the xy-plane and above by the graph z=f(x)=xln(xy)

Answer 1

I believe you can set it up as
$$
\Large \int_{1}^{e} \int_{1}^{2} x\ln(xy) ~dx ~ dy
$$

Answer 2

Yeah I understand that, I just can't quite get the integration right

Answer 3

$$\Large \int_{1}^{e} \int_{1}^{2} x\ln(xy) ~dx ~ dy\\~\\
\Large \int_{1}^{e} \left( \int_{1}^{2} x\ln(xy) ~dx \right ) ~ dy\\~\\
\\\Large = \int_{1}^{e} \left[ \frac 12 x^2~\ln(xy)-\frac14x^2 \right]_{1}^{2}~ dy\\~\\
\\\Large = \int_{1}^{e} (3/2)\ln(y)-3/4+2~\ln(2)~ dy\\~\\
\Large =\left[ (3/2)y\ln(y)-(9/4)y+2\ln(2)y \right]_{1}^{e}\\~\\
\Large=\left[ (3/2)y\ln(y)-(9/4)y+2\ln(2)y \right]_{1}^{e}\\~\\
\Large = 9/4-2\ln(2)-(3/4)e+2\ln(2)e\\~\\
\\\Large \approx 2.593333

$$