\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(14e^{(2x^{2} + 2y^{2})} -\frac{ (\frac{5}{3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})}{17})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(14\cdot re^{(2\cdot r^{2})} -\frac{ (5\cdot r)}{(17\cdot 3^{(\frac{(3\cdot r^{2})}{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.685}{-0.729})=0.76\pi;\theta_2=arctan(\frac{-0.790}{0.613})=1.71\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.76\pi}^{1.71\pi}\int_{0.5}^{1.93}(14\cdot re^{(2\cdot r^{2})} -\frac{ (5\cdot r)}{(17\cdot 3^{(\frac{(3\cdot r^{2})}{2})})})drd\theta=(\frac{(7e^{(2\cdot r^{2})})}{2}+\frac{ 5}{(51\cdot 3^{(\frac{(3\cdot r^{2})}{2})}ln(3))})d\theta|_{0.5}^{1.93}\\&\int_{0.76\pi}^{1.71\pi}(6012.0)d\theta=(6012.0\theta)|_{0.76\pi}^{1.71\pi}=17944.3\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(19\cdot (\frac{1}{2})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{3}-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{2})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot (\frac{1}{2})^{(\frac{r^{2}}{2})}\cdot r)}{3}-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{2})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.509}{-0.861})=0.83\pi;\theta_2=arctan(\frac{-0.876}{-0.482})=1.34\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.83\pi}^{1.34\pi}\int_{0}^{1.6}(\frac{(19\cdot (\frac{1}{2})^{(\frac{r^{2}}{2})}\cdot r)}{3}-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{2})drd\theta=(-\frac{ sin(\frac{(3\cdot r^{2})}{2})}{2}-\frac{ 19}{(3\cdot 2^{(\frac{r^{2}}{2})}ln(2))})d\theta|_{0}^{1.6}\\&\int_{0.83\pi}^{1.34\pi}(5.696)d\theta=(5.696\theta)|_{0.83\pi}^{1.34\pi}=9.13\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (17e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{2}-\frac{ 31}{(5\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (17\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{2}-\frac{ 93}{(10\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.661}{0.750})=0.23\pi;\theta_2=arctan(\frac{-0.249}{0.969})=1.92\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.23\pi}^{1.92\pi}\int_{0.49}^{1.43}(-\frac{ (17\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{2}-\frac{ 93}{(10\cdot r)})drd\theta=(\frac{(17e^{(-\frac{(3\cdot r^{2})}{2})})}{6}-\frac{ (93ln(r))}{10})d\theta|_{0.49}^{1.43}\\&\int_{0.23\pi}^{1.92\pi}(-11.81)d\theta=(-11.81\theta)|_{0.23\pi}^{1.92\pi}=-62.68\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(x^{2} + y^{2})}{41}-\frac{ (38e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(r^{2}))}{41}-\frac{ (38\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.2\pi}^{1.78\pi}\int_{0.2}^{1.32}(\frac{(rsin(r^{2}))}{41}-\frac{ (38\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{5})drd\theta=(-\frac{ cos(r^{2})}{82}-\frac{ (57e^{(\frac{(2\cdot r^{2})}{3})})}{10})d\theta|_{0.2}^{1.32}\\&\int_{0.2\pi}^{1.78\pi}(-12.34)d\theta=(-12.34\theta)|_{0.2\pi}^{1.78\pi}=-61.27\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{22}{(5\cdot (\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}-\frac{ 16}{(3\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{236}{(45\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.094}{-0.996})=0.97\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.97\pi}^{1.51\pi}\int_{0.31}^{0.91}(\frac{236}{(45\cdot r)})drd\theta=(\frac{(236ln(r))}{45})d\theta|_{0.31}^{0.91}\\&\int_{0.97\pi}^{1.51\pi}(5.648)d\theta=(5.648\theta)|_{0.97\pi}^{1.51\pi}=9.58\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4sin(x^{2} + y^{2}))}{21}+\frac{ 47}{(5\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot rsin(r^{2}))}{21}+\frac{ 94}{(15\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.063}{-0.998})=0.98\pi;\theta_2=arctan(\frac{-0.536}{0.844})=1.82\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.98\pi}^{1.82\pi}\int_{0.26}^{1.54}(\frac{(4\cdot rsin(r^{2}))}{21}+\frac{ 94}{(15\cdot r)})drd\theta=(\frac{(94ln(r))}{15}-\frac{ (2cos(r^{2}))}{21})d\theta|_{0.26}^{1.54}\\&\int_{0.98\pi}^{1.82\pi}(11.31)d\theta=(11.31\theta)|_{0.98\pi}^{1.82\pi}=29.85\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(3cos(2x^{2} + 2y^{2}) +\frac{ (35e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(3\cdot rcos(2\cdot r^{2}) +\frac{ (35\cdot re^{(-\frac{r^{2}}{2})})}{4})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.707}{0.707})=1.75\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.75\pi}^{1.75\pi}\int_{0.2}^{1.96}(3\cdot rcos(2\cdot r^{2}) +\frac{ (35\cdot re^{(-\frac{r^{2}}{2})})}{4})drd\theta=(\frac{(3sin(2\cdot r^{2}))}{4}-\frac{ (35e^{(-\frac{r^{2}}{2})})}{4})d\theta|_{0.2}^{1.96}\\&\int_{0.75\pi}^{1.75\pi}(7.974)d\theta=(7.974\theta)|_{0.75\pi}^{1.75\pi}=25.05\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(2e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})} +\frac{ sin(x^{2} + y^{2})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(r^{2}))}{3}+ 2\cdot re^{(-\frac{(2\cdot r^{2})}{3})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.339}{0.941})=0.11\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.11\pi}^{1.97\pi}\int_{0.12}^{1.26}(\frac{(rsin(r^{2}))}{3}+ 2\cdot re^{(-\frac{(2\cdot r^{2})}{3})})drd\theta=(-\frac{ cos(r^{2})}{6}-\frac{ (3e^{(-\frac{(2\cdot r^{2})}{3})})}{2})d\theta|_{0.12}^{1.26}\\&\int_{0.11\pi}^{1.97\pi}(1.135)d\theta=(1.135\theta)|_{0.11\pi}^{1.97\pi}=6.63\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{21}- 13cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{21}- 13\cdot rcos(\frac{(3\cdot r^{2})}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.125}{-0.992})=0.96\pi;\theta_2=arctan(\frac{-0.339}{0.941})=1.89\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.96\pi}^{1.89\pi}\int_{0}^{1}(\frac{(2\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{21}- 13\cdot rcos(\frac{(3\cdot r^{2})}{2}))drd\theta=(-\frac{ (2e^{(-\frac{(3\cdot r^{2})}{2})})}{63}-\frac{ (13sin(\frac{(3\cdot r^{2})}{2}))}{3})d\theta|_{0}^{1}\\&\int_{0.96\pi}^{1.89\pi}(-4.298)d\theta=(-4.298\theta)|_{0.96\pi}^{1.89\pi}=-12.56\end{align*}\)

\(\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{4}{(41\cdot (x^{2} + y^{2}))}-\frac{ 1}{(5\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{22}{(615\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}\)

\(\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.218}{-0.976})=0.93\pi;\theta_2=arctan(\frac{-0.992}{-0.125})=1.46\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.93\pi}^{1.46\pi}\int_{0.36}^{1.6}(-\frac{22}{(615\cdot r)})drd\theta=(-\frac{(22ln(r))}{615})d\theta|_{0.36}^{1.6}\\&\int_{0.93\pi}^{1.46\pi}(-0.05336)d\theta=(-0.05336\theta)|_{0.93\pi}^{1.46\pi}=-0.09\end{align*}\)