$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(29\cdot 3^{(x^{2} + y^{2})}sin(z + 1))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(29\cdot 3^{(r^{2})}\cdot rsin(z + 1))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.951}{0.309})=0.4\pi;\theta_2=arctan(\frac{-0.827}{-0.562})=1.31\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.63}\int_{0.4\pi}^{1.31\pi}\int_{0}^{1.15}(29\cdot 3^{(r^{2})}\cdot rsin(z + 1))drd\theta dz=(\frac{(29\cdot 3^{(r^{2})}sin(z + 1))}{(2ln(3))})d\theta dz|_{0}^{1.15}\\&\int_{0}^{0.63}\int_{0.4\pi}^{1.31\pi}(43.23sin(z + 1))d\theta dz=(43.23\theta sin(z + 1))dz|_{0.4\pi}^{1.31\pi}\\&\int_{0}^{0.63}(123.6sin(z + 1))dz=(-123.6cos(z + 1))|_{0}^{0.63}=74.09\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}\cdot (z + 1)^{2})}{41})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})}\cdot (z + 1)^{2})}{41})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.930}{-0.368})=0.62\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{0.88}\int_{0.62\pi}^{1.65\pi}\int_{0}^{0.63}(-\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})}\cdot (z + 1)^{2})}{41})drd\theta dz=(\frac{(9e^{(-\frac{(2\cdot r^{2})}{3})}\cdot (z + 1)^{2})}{164})d\theta dz|_{0}^{0.63}\\&\int_{0}^{0.88}\int_{0.62\pi}^{1.65\pi}(-0.01276\cdot (z + 1)^{2})d\theta dz=(-0.01276\theta \cdot (z + 1)^{2})dz|_{0.62\pi}^{1.65\pi}\\&\int_{0}^{0.88}(-0.04128\cdot (z + 1)^{2})dz=(-0.01376\cdot (z + 1)^{3})|_{0}^{0.88}=-0.08\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(cos(2x^{2} + 2y^{2})e^{(2z)})}{8})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(re^{(2z)}cos(2\cdot r^{2}))}{8})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.426}{-0.905})=0.86\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.63}\int_{0.86\pi}^{1.87\pi}\int_{0}^{1.18}(-\frac{(re^{(2z)}cos(2\cdot r^{2}))}{8})drd\theta dz=(-\frac{(e^{(2z)}sin(2\cdot r^{2}))}{32})d\theta dz|_{0}^{1.18}\\&\int_{0}^{1.63}\int_{0.86\pi}^{1.87\pi}(-0.01091e^{(2z)})d\theta dz=(-0.01091\theta e^{(2z)})dz|_{0.86\pi}^{1.87\pi}\\&\int_{0}^{1.63}(-0.03463e^{(2z)})dz=(-0.01732e^{(2z)})|_{0}^{1.63}=-0.43\end{align*}$

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

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.960}{-0.279})=0.59\pi;\theta_2=arctan(\frac{-0.891}{-0.454})=1.35\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0}^{1.09}\int_{0.59\pi}^{1.35\pi}\int_{0.17}^{1.26}(\frac{29}{r})drd\theta dz=(29ln(r))d\theta dz|_{0.17}^{1.26}\\&\int_{0}^{1.09}\int_{0.59\pi}^{1.35\pi}(58.09)d\theta dz=(58.09\theta)dz|_{0.59\pi}^{1.35\pi}\\&\int_{0}^{1.09}(138.7)dz=(138.7z)|_{0}^{1.09}=151.18\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(\frac{19}{(\frac{7}{2})^{(2x^{2} + 2y^{2})}})}{(3\cdot 2^z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot r)}{(3\cdot 2^z\cdot (\frac{7}{2})^{(2\cdot r^{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.988}{0.156})=1.55\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{0.99}\int_{0.79\pi}^{1.55\pi}\int_{0.39}^{1.86}(\frac{(19\cdot r)}{(3\cdot 2^z\cdot (\frac{7}{2})^{(2\cdot r^{2})})})drd\theta dz=(-\frac{(19\cdot (\frac{4}{49})^{(r^{2})}\cdot (\frac{1}{2})^z)}{(12ln(\frac{7}{2}))})d\theta dz|_{0.39}^{1.86}\\&\int_{0}^{0.99}\int_{0.79\pi}^{1.55\pi}(\frac{0.8632}{2^{(z)}})d\theta dz=(\frac{(0.8632\theta )}{2^{(z)}})dz|_{0.79\pi}^{1.55\pi}\\&\int_{0}^{0.99}(\frac{2.061}{2^{(z)}})dz=(-\frac{2.973}{2^{(z)}})|_{0}^{0.99}=1.48\end{align*}$

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

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.960}{-0.279})=0.59\pi;\theta_2=arctan(\frac{-0.249}{0.969})=1.92\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0}^{1.63}\int_{0.59\pi}^{1.92\pi}\int_{0}^{1.15}(rcos(\frac{(2\cdot r^{2})}{3})\cdot (\frac{(3z)}{13}+\frac{ 3}{13}))drd\theta dz=(sin(\frac{(2\cdot r^{2})}{3})\cdot (\frac{(9z)}{52}+\frac{ 9}{52}))d\theta dz|_{0}^{1.15}\\&\int_{0}^{1.63}\int_{0.59\pi}^{1.92\pi}(0.1336z + 0.1336)d\theta dz=(0.1336\theta \cdot (z + 1))dz|_{0.59\pi}^{1.92\pi}\\&\int_{0}^{1.63}(0.5581z + 0.5581)dz=(0.2791\cdot (z + 1)^{2})|_{0}^{1.63}=1.65\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(3cos(2x^{2} + 2y^{2})e^{(-2z)})}{28})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(3\cdot re^{(-2z)}cos(2\cdot r^{2}))}{28})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.982}{-0.187})=0.56\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.95}\int_{0.56\pi}^{1.77\pi}\int_{0.49}^{1.53}(-\frac{(3\cdot re^{(-2z)}cos(2\cdot r^{2}))}{28})drd\theta dz=(-\frac{(3e^{(-2z)}sin(2\cdot r^{2}))}{112})d\theta dz|_{0.49}^{1.53}\\&\int_{0}^{1.95}\int_{0.56\pi}^{1.77\pi}(0.03915e^{(-2z)})d\theta dz=(0.03915\theta e^{(-2z)})dz|_{0.56\pi}^{1.77\pi}\\&\int_{0}^{1.95}(0.1488e^{(-2z)})dz=(-0.07441e^{(-2z)})|_{0}^{1.95}=0.07\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(44\cdot 3^{(\frac{z}{2})}e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(44\cdot 3^{(\frac{z}{2})}\cdot re^{(\frac{(3\cdot r^{2})}{2})})}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{0.156}{-0.988})=0.95\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.21}\int_{0.2\pi}^{0.95\pi}\int_{0}^{1.82}(\frac{(44\cdot 3^{(\frac{z}{2})}\cdot re^{(\frac{(3\cdot r^{2})}{2})})}{5})drd\theta dz=(\frac{(44\cdot 3^{(\frac{z}{2})}e^{(\frac{(3\cdot r^{2})}{2})})}{15})d\theta dz|_{0}^{1.82}\\&\int_{0}^{1.21}\int_{0.2\pi}^{0.95\pi}(419\cdot 3^{(0.5z)})d\theta dz=(419\cdot 3^{(0.5z)}\theta)dz|_{0.2\pi}^{0.95\pi}\\&\int_{0}^{1.21}(987.1\cdot 3^{(0.5z)})dz=(1797\cdot 3^{(0.5z)})|_{0}^{1.21}=1696.12\end{align*}$

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

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.729}{-0.685})=0.74\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.6}\int_{0.74\pi}^{1.97\pi}\int_{0.32}^{1.19}(-rcos(r^{2})e^{(2z)})drd\theta dz=(-\frac{(sin(r^{2})e^{(2z)})}{2})d\theta dz|_{0.32}^{1.19}\\&\int_{0}^{0.6}\int_{0.74\pi}^{1.97\pi}(-0.4429e^{(2z)})d\theta dz=(-0.4429\theta e^{(2z)})dz|_{0.74\pi}^{1.97\pi}\\&\int_{0}^{0.6}(-1.712e^{(2z)})dz=(-0.8558e^{(2z)})|_{0}^{0.6}=-1.99\end{align*}$

$\displaystyle \begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(27e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}\cdot (z + 1)^{2})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(27\cdot re^{(-\frac{(3\cdot r^{2})}{2})}\cdot (z + 1)^{2})}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\displaystyle \begin{align*}&\theta_1=arctan(\frac{0.125}{0.992})=0.04\pi;\theta_2=arctan(\frac{0.941}{-0.339})=0.61\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{0.86}\int_{0.04\pi}^{0.61\pi}\int_{0}^{1.39}(\frac{(27\cdot re^{(-\frac{(3\cdot r^{2})}{2})}\cdot (z + 1)^{2})}{5})drd\theta dz=(-\frac{(9e^{(-\frac{(3\cdot r^{2})}{2})}\cdot (z + 1)^{2})}{5})d\theta dz|_{0}^{1.39}\\&\int_{0}^{0.86}\int_{0.04\pi}^{0.61\pi}(1.701\cdot (z + 1)^{2})d\theta dz=(1.701\theta \cdot (z + 1)^{2})dz|_{0.04\pi}^{0.61\pi}\\&\int_{0}^{0.86}(3.046\cdot (z + 1)^{2})dz=(1.015\cdot (z + 1)^{3})|_{0}^{0.86}=5.52\end{align*}$