Question

Find the volume of the solid bounded below by the rectangle R : 1≤x≤e, 1≤y≤2 in the xy -plane and above by the graph of z=f(x,y)=ln(x)y .

Answer 1

please show ur attempt

Answer 2

\[\int\limits\limits_{1}^{2}\int\limits\limits_{1}^{e}\frac{ \ln(x) }{ y }dxdy=\int\limits\limits_{1}^{2}\frac{ e }{ y }\ln(e)-e-\frac{ 1 }{ y }\ln(y)+1dy\]
\[\int\limits_{1}^{2}\frac{ e }{ y }-e+1=eln(2)-2e+2-eln(1)-e+1\]
\[\approx0.165887556904675\]

Answer 3

what is the integral of ln(x) ?

Answer 4

xln(x)-x

Answer 5

yeah, that is what i was trying to remember off the top of my head :)

Answer 6

xlnx - x evaluated at 1 and e

elne - e - (1ln1 - 1)

e -e +1 = 1

Answer 7

so determine the integral of 1/y from 1 to 2

Answer 8

I got it right the answer is ln(2), thanks!

Answer 9

\[\int\limits\limits_{1}^{2}\int\limits\limits_{1}^{e}\frac{ \ln(x) }{ y }dxdy=\int_{1}^{2}\frac 1y\left[\int\limits\limits_{1}^{e}ln(x) dx\right]dy\]

Answer 10

ln2 is good for me as well :)