Question

Let D={(x,y,z)∈R^3:1≤x≤3,0≤y≤ln(z),0≤z≤x}. The answer to ∭D e^yln(x)dV

is
Select one:
a. 13/9
b. 3/2
c. 5/9
d. -13/9
e. 2
f. 1/2

Answer 1

$\begin{aligned} \iiint_De^y \ln(x)dv&=\int_{1}^{3} \int_{0}^{x} ( \int_{0}^{\ln z} e^y\ln(x) dy)dzdx\\ &= \int_{1}^{3} \int_{0}^{x} ( \ln(x) e^y \bigg| _{0}^{\ln z} )dzdx\\ &= \int_{1}^{3} \int_{0}^{x} ( \ln(x) (e^{\ln z}-1) )dzdx\\ &= \int_{1}^{3} \int_{0}^{x} ( \ln(x) (z-1)) )dzdx\\ &= \int_{1}^{3} (\ln(x) (\frac {1}{2}z^2-z) \bigg| _{0}^{x} ) dx\\ &= \int_{1}^{3} (\ln(x) (\frac {1}{2}x^2-x) ) dx\\ &= \int_{1}^{3}\frac {1}{2}x^2\ln(x) dx-\int_{1}^{3} x\ln(x) dx \end{aligned}\\ \text {To evaluate \displaystyle } \int_{1}^{3}\frac {1}{2}x^2\ln(x) dx \\ \text {use integration by parts , we let } \\ \begin{aligned} u&=\ln x\ \ \ \ \ \ \ \ \ \ &dv&= x^2dx\\ du&= \frac{1}{x} \ \ \ \ \ \ \ & v&= \frac {1}{3}x^3 \end{aligned}\\ \begin{aligned} \int_{1}^{3}\frac {1}{2}x^2\ln(x) dx&= \frac{1}{2}\left( \frac {1}{3}x^3\ln x\bigg|_{1}^{3}- \int_{1}^{3}\frac {1}{3}x^2dx\right)\\ &=\frac{1}{2}\left (\frac {1}{3}x^3\ln(x)-\frac {1}{9}x^3\right) \bigg| _{1}^{3}\\ &= \frac {9}{2}\ln 3-\frac {13}{9} \end{aligned}\\ \text {In the same way, we get } \int_{1}^{3} x\ln(x) dx =-2+\frac {9}{2}\ln 3\\ \text{Hence} \\ \begin{aligned} \int_{1}^{3} \ln(x) (\frac {1}{2}x^2-x) dx &=\frac {9}{2}\ln 3-\frac {13}{9}-(-2+\frac {9}{2}\ln 3)\\ &=\frac {5}{9} \end{aligned}$