$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(5cos(x^{2} + y^{2}) +\frac{ (3\cdot 2^ze^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{13})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(5\cdot rcos(r^{2}) +\frac{ (3\cdot 2^z\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{13})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.156}{0.988})=0.05\pi;\theta_2=arctan(\frac{0.031}{-1.000})=0.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.31}\int_{0.05\pi}^{0.99\pi}\int_{0.31}^{1.89}(5\cdot rcos(r^{2}) +\frac{ (3\cdot 2^z\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{13})drd\theta dz=(\frac{(5sin(r^{2}))}{2}-\frac{ (2^ze^{(-\frac{(3\cdot r^{2})}{2})})}{13})d\theta dz|_{0.31}^{1.89}\\&\int_{0}^{1.31}\int_{0.05\pi}^{0.99\pi}(0.06623\cdot 2^z - 1.283)d\theta dz=(\theta\cdot (0.06623\cdot 2^z - 1.283))dz|_{0.05\pi}^{0.99\pi}\\&\int_{0}^{1.31}(0.1956\cdot 2^z - 3.789)dz=(0.2822\cdot 2^z - 3.789z)|_{0}^{1.31}=-4.55\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- 16e^{(x^{2} + y^{2})} -\frac{ (5e^{(2z)})}{(19\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(- 16\cdot re^{(r^{2})} -\frac{ (5e^{(2z)})}{(19\cdot r)})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.930}{-0.368})=0.62\pi;\theta_2=arctan(\frac{-0.988}{-0.156})=1.45\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.72}\int_{0.62\pi}^{1.45\pi}\int_{0.42}^{1.42}(- 16\cdot re^{(r^{2})} -\frac{ (5e^{(2z)})}{(19\cdot r)})drd\theta dz=(- 8e^{(r^{2})} -\frac{ (5e^{(2z)}ln(r))}{19})d\theta dz|_{0.42}^{1.42}\\&\int_{0}^{1.72}\int_{0.62\pi}^{1.45\pi}(- 0.3206e^{(2z)} - 50.55)d\theta dz=(-1.0\theta\cdot (0.3206e^{(2z)} + 50.55))dz|_{0.62\pi}^{1.45\pi}\\&\int_{0}^{1.72}(- 0.8359e^{(2z)} - 131.8)dz=(- 131.8z - 0.4179e^{(2z)})|_{0}^{1.72}=-239.31\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2\cdot 4^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{7}-\frac{ (7\cdot 3^zcos(x^{2} + y^{2}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot 4^{(\frac{r^{2}}{2})}\cdot r)}{7}-\frac{ (7\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.09}\int_{0.16\pi}^{1.62\pi}\int_{0.27}^{1.86}(\frac{(2\cdot 4^{(\frac{r^{2}}{2})}\cdot r)}{7}-\frac{ (7\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz=(\frac{2^{(r^{2})}}{(7ln(2))}-\frac{ (7\cdot 3^zsin(r^{2}))}{4})d\theta dz|_{0.27}^{1.86}\\&\int_{0}^{1.09}\int_{0.16\pi}^{1.62\pi}(0.6746\cdot 3^z + 2.051)d\theta dz=(\theta\cdot (0.6746\cdot 3^z + 2.051))dz|_{0.16\pi}^{1.62\pi}\\&\int_{0}^{1.09}(3.094\cdot 3^z + 9.405)dz=(9.405z + 2.817\cdot 3^z)|_{0}^{1.09}=16.76\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(- x^{2} - y^{2})})}{11}+\frac{ (49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-r^{2})})}{11}+\frac{ (49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.187}{0.982})=1.94\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}\int_{0}^{1.87}(\frac{(3\cdot re^{(-r^{2})})}{11}+\frac{ (49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz=(\frac{(49sin(r^{2})e^{(2z)})}{12}-\frac{ (3e^{(-r^{2})})}{22})d\theta dz|_{0}^{1.87}\\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}(0.1322 - 1.421e^{(2z)})d\theta dz=(-\theta\cdot (1.421e^{(2z)} - 0.1322))dz|_{0.82\pi}^{1.94\pi}\\&\int_{0}^{1.45}(0.4653 - 4.998e^{(2z)})dz=(0.4653z - 2.499e^{(2z)})|_{0}^{1.45}=-42.24\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(48cos(3z)sin(2x^{2} + 2y^{2}) - 20sin(x^{2} + y^{2}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(48\cdot rcos(3z)sin(2\cdot r^{2}) - 20\cdot rsin(r^{2}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.992}{0.125})=1.54\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{0.54}\int_{0.75\pi}^{1.54\pi}\int_{0.17}^{1.61}(48\cdot rcos(3z)sin(2\cdot r^{2}) - 20\cdot rsin(r^{2}))drd\theta dz=(10cos(r^{2}) - 12cos(3z)cos(2\cdot r^{2}))d\theta dz|_{0.17}^{1.61}\\&\int_{0}^{0.54}\int_{0.75\pi}^{1.54\pi}(6.526cos(3z) - 18.52)d\theta dz=(\theta\cdot (6.526cos(3z) - 18.52))dz|_{0.75\pi}^{1.54\pi}\\&\int_{0}^{0.54}(16.2cos(3z) - 45.97)dz=(10.8cos(1.5z)sin(1.5z) - 45.97z)|_{0}^{0.54}=-19.43\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(13cos(2x^{2} + 2y^{2}))}{6}-\frac{ (2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{15})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(13\cdot rcos(2\cdot r^{2}))}{6}-\frac{ (2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{0.536}{-0.844})=0.82\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.01}\int_{0.16\pi}^{0.82\pi}\int_{0}^{1.94}(\frac{(13\cdot rcos(2\cdot r^{2}))}{6}-\frac{ (2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz=(\frac{(13sin(2\cdot r^{2}))}{24}-\frac{ (2e^{(\frac{(3\cdot r^{2})}{2}- 2z)})}{45})d\theta dz|_{0}^{1.94}\\&\int_{0}^{1.01}\int_{0.16\pi}^{0.82\pi}(0.513 - 12.53e^{(-2z)})d\theta dz=(-\theta\cdot (12.53e^{(-2z)} - 0.513))dz|_{0.16\pi}^{0.82\pi}\\&\int_{0}^{1.01}(1.064 - 25.99e^{(-2z)})dz=(1.064z + 12.99e^{(-2z)})|_{0}^{1.01}=-10.2\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{29}{(3\cdot (2x^{2} + 2y^{2}))}-\frac{ (4sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-z)})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{29}{(6\cdot r)}-\frac{ (4\cdot re^{(-z)}sin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.861}{-0.509})=1.33\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.88}\int_{0.47\pi}^{1.33\pi}\int_{0.33}^{1.47}(\frac{29}{(6\cdot r)}-\frac{ (4\cdot re^{(-z)}sin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta dz=(\frac{(29ln(r))}{6}+\frac{ (4e^{(-z)}cos(\frac{(3\cdot r^{2})}{2}))}{9})d\theta dz|_{0.33}^{1.47}\\&\int_{0}^{1.88}\int_{0.47\pi}^{1.33\pi}(7.221 - 0.8808e^{(-z)})d\theta dz=(-\theta\cdot (0.8808e^{(-z)} - 7.221))dz|_{0.47\pi}^{1.33\pi}\\&\int_{0}^{1.88}(19.51 - 2.38e^{(-z)})dz=(19.51z + 2.38e^{(-z)})|_{0}^{1.88}=34.66\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{5}{(39\cdot (x^{2} + y^{2}))}+\frac{ (4sin(2x^{2} + 2y^{2})e^{(-2z)})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{5}{(39\cdot r)}+\frac{ (4\cdot re^{(-2z)}sin(2\cdot r^{2}))}{3})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.62}\int_{0.26\pi}^{1.57\pi}\int_{0.4}^{1.4}(\frac{5}{(39\cdot r)}+\frac{ (4\cdot re^{(-2z)}sin(2\cdot r^{2}))}{3})drd\theta dz=(\frac{(5ln(r))}{39}-\frac{ (2cos(r^{2})^{2}e^{(-2z)})}{3})d\theta dz|_{0.4}^{1.4}\\&\int_{0}^{0.62}\int_{0.26\pi}^{1.57\pi}(0.5538e^{(-2z)} + 0.1606)d\theta dz=(\theta\cdot (0.5538e^{(-2z)} + 0.1606))dz|_{0.26\pi}^{1.57\pi}\\&\int_{0}^{0.62}(2.279e^{(-2z)} + 0.661)dz=(0.661z - 1.139e^{(-2z)})|_{0}^{0.62}=1.22\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(4cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}) - 21cos(z + 2)e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(4\cdot rcos(\frac{(3\cdot r^{2})}{2}) - 21\cdot rcos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{0.187}{-0.982})=0.94\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.37}\int_{0.26\pi}^{0.94\pi}\int_{0}^{1.81}(4\cdot rcos(\frac{(3\cdot r^{2})}{2}) - 21\cdot rcos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})drd\theta dz=(\frac{(4sin(\frac{(3\cdot r^{2})}{2}))}{3}-\frac{ (63cos(z + 2)e^{(\frac{(2\cdot r^{2})}{3})})}{4})d\theta dz|_{0}^{1.81}\\&\int_{0}^{1.37}\int_{0.26\pi}^{0.94\pi}(- 124.1cos(z + 2) - 1.306)d\theta dz=(-\theta\cdot (124.1cos(z + 2) + 1.306))dz|_{0.26\pi}^{0.94\pi}\\&\int_{0}^{1.37}(- 265.2cos(z + 2) - 2.791)dz=(- 2.791z - 265.2sin(z + 2))|_{0}^{1.37}=297.39\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4cos(x^{2} + y^{2}))}{9}+\frac{ (cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})e^{(-z)})}{4})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot rcos(r^{2}))}{9}+\frac{ (re^{(-z)}cos(\frac{(2\cdot r^{2})}{3}))}{4})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.339}{0.941})=0.11\pi;\theta_2=arctan(\frac{-0.809}{0.588})=1.7\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.1}\int_{0.11\pi}^{1.7\pi}\int_{0}^{0.86}(\frac{(4\cdot rcos(r^{2}))}{9}+\frac{ (re^{(-z)}cos(\frac{(2\cdot r^{2})}{3}))}{4})drd\theta dz=(\frac{(2sin(r^{2}))}{9}+\frac{ (3e^{(-z)}sin(\frac{(2\cdot r^{2})}{3}))}{16})d\theta dz|_{0}^{0.86}\\&\int_{0}^{1.1}\int_{0.11\pi}^{1.7\pi}(0.08875e^{(-z)} + 0.1498)d\theta dz=(\theta\cdot (0.08875e^{(-z)} + 0.1498))dz|_{0.11\pi}^{1.7\pi}\\&\int_{0}^{1.1}(0.4433e^{(-z)} + 0.7482)dz=(0.7482z - 0.4433e^{(-z)})|_{0}^{1.1}=1.12\end{align*}$