Question

2.Sketch the region whose area is given by the integral and evaluate the integral.

Answer 1

2. The integration region,

$\left\{(r,\theta)\mid\dfrac\pi6\le\theta\le\dfrac\pi2\land2\le r\le3\right\}$

corresponds to what you might call an "annular sector" (i.e. the analog of circular sector for the annulus or ring). In other words, it's the region between the two circles of radii $r=2$ and $r=3$, taken between the rays $\theta=\frac\pi6$ and $\theta=\frac\pi2$. (The previous question of yours that I just posted an answer to has a similar region with slightly different parameters.)

You can separate the variables to compute the integral:

$\displaystyle\int_{\pi/6}^{\pi/2}\int_2^3r^2\sin^2\theta\,\mathrm dr\,\mathrm d\theta=\left(\int_{\pi/6}^{\pi/2}\sin^2\theta\,\mathrm d\theta\right)\left(\int_2^3r^2\,\mathrm dr\right)$

which should be doable for you. You would find it has a value of 19/72*(3√3 + 4π).

3. Without knowing the definition of the region D, the best we can do is convert what we can to polar coordinates. Namely,

$r^2=x^2+y^2$

so that

$\displaystyle\iint_De^{x^2+y^2}\,\mathrm dA=\iint_Dre^{r^2}\,\mathrm dr\,\mathrm d\theta$