Triple Integration in Cylindrical Coordinates

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AP Calculus BC › Triple Integration in Cylindrical Coordinates

Questions 1 - 10

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(e^{(- x^{2} - y^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.56\&\text{and length }1.4\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.187,0.982)\text{ and }\overrightarrow{u_2}=(-0.729,-0.685)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(e^{(- x^{2} - y^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(re^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.982}{-0.187})=0.56\pi;\theta_2=arctan(\frac{-0.685}{-0.729})=1.24\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0}^{1.4}\int_{0.56\pi}^{1.24\pi}\int_{0}^{1.56}(re^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))drd\theta dz=(-\frac{(e^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))}{2})d\theta dz|_{0}^{1.56}\&\int_{0}^{1.4}\int_{0.56\pi}^{1.24\pi}(0.2851z + 0.2851)d\theta dz=(0.2851\theta \cdot (z + 1))dz|_{0.56\pi}^{1.24\pi}\&\int_{0}^{1.4}(0.609z + 0.609)dz=(0.3045\cdot (z + 1)^{2})|_{0}^{1.4}=1.45\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.87\&\text{and length }1.45\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.844,0.536)\text{ and }\overrightarrow{u_2}=(0.982,-0.187)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

$\begin{align*}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.187}{0.982})=1.94\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}\int_{0}^{1.87}(\frac{(49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz=(\frac{(49sin(r^{2})e^{(2z)})}{12})d\theta dz|_{0}^{1.87}\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}(-1.421e^{(2z)})d\theta dz=(-1.421\theta e^{(2z)})dz|_{0.82\pi}^{1.94\pi}\&\int_{0}^{1.45}(-4.998e^{(2z)})dz=(-2.499e^{(2z)})|_{0}^{1.45}=-42.92\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{42}{2^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}\cdot 2^z})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin,}\&\text{with inner/outer radii }\&0.43\text{ and }1.71\&\text{and length }0.72\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.063,0.998)\text{ and }\overrightarrow{u_2}=(-0.941,-0.339)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{42}{2^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}\cdot 2^z})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(42\cdot 2^z\cdot r)}{2^{(\frac{(2\cdot r^{2})}{3})}})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.998}{-0.063})=0.52\pi;\theta_2=arctan(\frac{-0.339}{-0.941})=1.11\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\&\int_{0}^{0.72}\int_{0.52\pi}^{1.11\pi}\int_{0.43}^{1.71}(\frac{(42\cdot 2^z\cdot r)}{2^{(\frac{(2\cdot r^{2})}{3})}})drd\theta dz=(-\frac{(63\cdot 2^{(z -\frac{ (2\cdot r^{2})}{3})})}{(2ln(2))})d\theta dz|_{0.43}^{1.71}\&\int_{0}^{0.72}\int_{0.52\pi}^{1.11\pi}(29.96\cdot 2^z)d\theta dz=(29.96\cdot 2^z\theta)dz|_{0.52\pi}^{1.11\pi}\&\int_{0}^{0.72}(55.53\cdot 2^z)dz=(80.11\cdot 2^z)|_{0}^{0.72}=51.84\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-10cos(z + 1)sin(2x^{2} + 2y^{2}))dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.22\&\text{and length }1.91\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.536,0.844)\text{ and }\overrightarrow{u_2}=(0.637,-0.771)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(-10cos(z + 1)sin(2x^{2} + 2y^{2}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-10\cdot rcos(z + 1)sin(2\cdot r^{2}))drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.844}{-0.536})=0.68\pi;\theta_2=arctan(\frac{-0.771}{0.637})=1.72\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[cos(az)]=\frac{sin(az)}{a}\&\int_{0}^{1.91}\int_{0.68\pi}^{1.72\pi}\int_{0}^{1.22}(-10\cdot rcos(z + 1)sin(2\cdot r^{2}))drd\theta dz=(\frac{(5cos(z + 1)cos(2\cdot r^{2}))}{2})d\theta dz|_{0}^{1.22}\&\int_{0}^{1.91}\int_{0.68\pi}^{1.72\pi}(-4.966cos(z + 1))d\theta dz=(-4.966\theta cos(z + 1))dz|_{0.68\pi}^{1.72\pi}\&\int_{0}^{1.91}(-16.23cos(z + 1))dz=(-16.23sin(z + 1))|_{0}^{1.91}=9.93\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(11cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})sin(3z))dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.31\&\text{and length }0.65\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(1.000,0.031)\text{ and }\overrightarrow{u_2}=(-0.976,-0.218)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

$\begin{align*}&\theta_1=arctan(\frac{0.031}{1.000})=0.01\pi;\theta_2=arctan(\frac{-0.218}{-0.976})=1.07\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[sin(az)]=-\frac{cos(az)}{a}\&\int_{0}^{0.65}\int_{0.01\pi}^{1.07\pi}\int_{0}^{1.31}(11\cdot rsin(3z)cos(\frac{(3\cdot r^{2})}{2}))drd\theta dz=(\frac{(11sin(3z)sin(\frac{(3\cdot r^{2})}{2}))}{3})d\theta dz|_{0}^{1.31}\&\int_{0}^{0.65}\int_{0.01\pi}^{1.07\pi}(1.971sin(3z))d\theta dz=(1.971\theta sin(3z))dz|_{0.01\pi}^{1.07\pi}\&\int_{0}^{0.65}(6.563sin(3z))dz=(-2.188cos(3z))|_{0}^{0.65}=3\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(6e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-z)})}{29})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin,}\&\text{with inner/outer radii }\&0.25\text{ and }1.23\&\text{and length }1.8\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.891,0.454)\text{ and }\overrightarrow{u_2}=(0.930,-0.368)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

$\begin{align*}&\theta_1=arctan(\frac{0.454}{-0.891})=0.85\pi;\theta_2=arctan(\frac{-0.368}{0.930})=1.88\pi\&\text{Now, utilizing integral rules:}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.8}\int_{0.85\pi}^{1.88\pi}\int_{0.25}^{1.23}(-\frac{(6\cdot re^{(-z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{29})drd\theta dz=(\frac{(9e^{(- z -\frac{ (2\cdot r^{2})}{3})})}{58})d\theta dz|_{0.25}^{1.23}\&\int_{0}^{1.8}\int_{0.85\pi}^{1.88\pi}(-0.09224e^{(-z)})d\theta dz=(-0.09224\theta e^{(-z)})dz|_{0.85\pi}^{1.88\pi}\&\int_{0}^{1.8}(-0.2985e^{(-z)})dz=(0.2985e^{(-z)})|_{0}^{1.8}=-0.25\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(44\cdot 3^{(\frac{z}{2})}e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{5})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.82\&\text{and length }1.21\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.988,0.156)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

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

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(e^{(2z)}sin(x^{2} + y^{2}))}{3})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&1.11\&\text{and length }1.19\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.426,0.905)\text{ and }\overrightarrow{u_2}=(-0.613,-0.790)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

$\begin{align*}&\theta_1=arctan(\frac{0.905}{-0.426})=0.64\pi;\theta_2=arctan(\frac{-0.790}{-0.613})=1.29\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[e^{az}]=\frac{e^{az}}{a}\&\int_{0}^{1.19}\int_{0.64\pi}^{1.29\pi}\int_{0}^{1.11}(-\frac{(rsin(r^{2})e^{(2z)})}{3})drd\theta dz=(\frac{(cos(r^{2})e^{(2z)})}{6})d\theta dz|_{0}^{1.11}\&\int_{0}^{1.19}\int_{0.64\pi}^{1.29\pi}(-0.1113e^{(2z)})d\theta dz=(-0.1113\theta e^{(2z)})dz|_{0.64\pi}^{1.29\pi}\&\int_{0}^{1.19}(-0.2273e^{(2z)})dz=(-0.1136e^{(2z)})|_{0}^{1.19}=-1.11\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(3e^{(- 2x^{2} - 2y^{2})} -\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})dA\text{, where D is}\&\text{the region of a cylinder centered on the origin with a radius of }\&0.84\&\text{and length }1.77\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

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

$\begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\&\int_{0}^{1.77}\int_{0.2\pi}^{1.09\pi}\int_{0}^{0.84}(3\cdot re^{(-2\cdot r^{2})} -\frac{ (rcos(\frac{r^{2}}{2}))}{(11\cdot 2^{(\frac{z}{2})})})drd\theta dz=(-\frac{ (3e^{(-2\cdot r^{2})})}{4}-\frac{ sin(\frac{r^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})d\theta dz|_{0}^{0.84}\&\int_{0}^{1.77}\int_{0.2\pi}^{1.09\pi}(0.5671 -\frac{ 0.03141}{2^{(0.5z)}})d\theta dz=(-1.0\theta\cdot (\frac{0.03141}{2^{(0.5z)}}- 0.5671))dz|_{0.2\pi}^{1.09\pi}\&\int_{0}^{1.77}(1.586 -\frac{ 0.08783}{2^{(0.5z)}})dz=(1.586z +\frac{ 0.2534}{2^{(0.5z)}})|_{0}^{1.77}=2.69\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ (sin(z + 1)cos(x^{2} + y^{2}))}{25})dA\text{, where D is}\&\text{the region of a cylindrical annulus centered on the origin with radii }\&0.23\text{ and }1.56\&\text{and length }1.97\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.094,0.996)\text{ and }\overrightarrow{u_2}=(0.637,-0.771)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ (sin(z + 1)cos(x^{2} + y^{2}))}{25})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (rcos(r^{2})sin(z + 1))}{25})drd\theta dz\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{-0.094})=0.53\pi;\theta_2=arctan(\frac{-0.771}{0.637})=1.72\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\end{align*}$

$\begin{align*}\&\int[sin(az)]=-\frac{cos(az)}{a}\&\int_{0}^{1.97}\int_{0.53\pi}^{1.72\pi}\int_{0.23}^{1.56}(-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (rcos(r^{2})sin(z + 1))}{25})drd\theta dz=(-\frac{ sin(\frac{(3\cdot r^{2})}{2})}{41}-\frac{ (sin(r^{2})sin(z + 1))}{50})d\theta dz|_{0.23}^{1.56}\&\int_{0}^{1.97}\int_{0.53\pi}^{1.72\pi}(0.01381 - 0.01195sin(z + 1))d\theta dz=(-1.0\theta\cdot (0.01195sin(z + 1) - 0.01381))dz|_{0.53\pi}^{1.72\pi}\&\int_{0}^{1.97}(0.05165 - 0.04467sin(z + 1))dz=0.03\end{align*}$

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