Question

calc iii integral question, just need to set this up

Answer 1

note x^2 +y^2 = b

and so we get

\[\int\limits_{1}^{e}(\ln b /\sqrt{b}) db = \int\limits_{1}^{e}(1/\sqrt{b})*\ln (b) db = 2\sqrt{b}*\ln (b) - \int\limits_{1}^{e}(1/b)*(2\sqrt{b})db = 2\sqrt{b}*\ln b -2*2\int\limits_{1}^{e}(1/2\sqrt{b})db = 2\sqrt{b}*\ln (b) -4\sqrt{b} +C\]

Answer 2

may be these above wrote a right way ?

Answer 3

\(\int\int_A \dfrac{\ln(x^2 + y^2)}{\sqrt{x^2 + y^2}} ~ dx ~ dx \qquad A: ~ 1 \le x^2 + y^2 \le e^5\)

Switch to polar: \(r^2 = x^2 + y^2\)

\(= \int_0^{2 \pi}\int_1^{e^{\frac{5}{2}}} \dfrac{\ln(r^2)}{\sqrt{r^2}} ~r ~ dr ~ d \theta\)

\(= 4 \pi \int_1^{e^{\frac{5}{2}}} \ln(r) ~ dr \)

By IBP:

\(\int \ln(r) ~ dr = r (\ln(r) - 1) + C\)

\(\implies 4 \pi \left(\frac{3}{2} e^{\frac{5}{2}} + 1 \right)\)

Answer 4

was able to figure it out but thank you both very much ~