Question

Find the area of the region that is bounded by the given curve and lies in the specified sector.
r = e−θ/12, π/2 ≤ θ ≤ π

Answer 1

that is e^(-pi/12)

Answer 2

Ok this is a mess and given how the last one went I will only take up one of your guesses... try [-6( e^(-pi/4) - e^(-5pi/24) - e^(-pi/6) + e^(-pi/8) )]

Answer 3

These freaking minus signs have to be what is screwing me up hold on submitting for a sec

Answer 4

I think it should be +6 times all that not -6

Answer 5

But that is only since the area is above the x-axis... my bounds all look correct so I am not sure why the math is telling me minus

Answer 6

I got that mess by doing:
\[\int f(r,\theta)rdrd\theta=(\int^{e^{-\pi/12}}_{e^{-\pi/24}}rdr)(\int^{\pi}_{\frac{\pi}{2}}e^{-\theta/12}d\theta)\]Seperated because the function doesn't depend on r so they are separable integrals

Answer 7

still here?

Answer 8

Ok I did a little digging and I figured out what was going wrong.... The function isnt e^-theta/12 that is the radius (face palms) so we need to be using the formula:
\[\frac{1}{2}\int r^2d\theta=\frac{1}{2}\int_{\pi/2}^{\pi} (e^{-\theta/12})^2d\theta=\frac{1}{2}\int_{\pi/2}^{\pi} e^{-\theta/6}d\theta\]

Answer 9

which gives:
\[-3(e^{-\pi/6}-e^{-\pi/12})\]Which is positive despite looking like it is negative

Answer 10

@alliguerrieri So I am assuming this would've worked on the last problem as well so sorry about that :( but try this method on the others.