Calculus 3

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AP Calculus BC › Calculus 3

Questions 1 - 10

$\begin{align*}&\text{Evaluate the triple integral}\int_{5}^{10}\int_{-2}^{3}\int_{1}^{5}(2x + 4y - 3x^2)dxdydz\end{align*}$

Explanation

$\begin{align*}&\text{When performing a triple integral, it is worth noting that the order}\&\text{in which the integration performed does not entirely matter, since each variable will be}\&\text{integrated independently, with the others being treated as constants for the}\&\text{puposes of integration. For example:}\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\&\text{Considering our integral:}\&\int_{5}^{10}\int_{-2}^{3}\int_{1}^{5}(2x + 4y - 3x^2)dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{5}^{10}\int_{-2}^{3}\int_{1}^{5}(2x + 4y - 3x^2)dxdydz=\int_{5}^{10}\int_{-2}^{3}(x{(x + 4y - x^2)})dydz|_{1}^{5}\&\int_{5}^{10}\int_{-2}^{3}(16y - 100)dydz=\int_{5}^{10}(4y{(2y - 25)})dz|_{-2}^{3}\&\int_{5}^{10}(-460)dz=-460zdz|_{5}^{10}=-2300\end{align*}$

Find the partial derivative of with the respect to .

Explanation

Assume that all other variables besides are constants.

Take the derivative of each term accordingly.

$\frac{\partial }{\partial x} (xy^2) = y^2$

$\frac{\partial }{\partial x} (xz) = z$

$\frac{\partial }{\partial x} (e^y) =0$

Sum all the terms.

The answer is:

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(10e^{(x^{2} + y^{2})} +\frac{ (\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.16\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.094,0.996)\text{ and }\overrightarrow{u_2}=(0.941,-0.339)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(10e^{(x^{2} + y^{2})} +\frac{ (\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(10\cdot re^{(r^{2})} +\frac{ (48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.339}{0.941})=1.89\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.47\pi}^{1.89\pi}\int_{0}^{1.16}(10\cdot re^{(r^{2})} +\frac{ (48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})})drd\theta=(5e^{(r^{2})} -\frac{ 36}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{7}))})d\theta|_{0}^{1.16}\&\int_{0.47\pi}^{1.89\pi}(26.14)d\theta=(26.14\theta)|_{0.47\pi}^{1.89\pi}=116.6\end{align*}$

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

Explanation

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

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

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(x^{4})}e^{(y^{2})}cos(z)\&\text{In the direction of }\overrightarrow{v}=(0,-4,-9).\end{align*}$

${0.91\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) - 0.82\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z)}$

${0.336\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z) - 0.828\cdot 2^{(x^{4})}e^{(y^{2})}sin(z)}$

${-54.6\cdot 2^{(x^{4})}x^{3}ye^{(y^{2})}sin(z)}$

${19.7\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z) - 9.85\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) + 27.3\cdot 2^{(x^{4})}x^{3}e^{(y^{2})}cos(z)}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-4)^2+(-9)^2}=9.85\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{9.85}=0\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4}{9.85}=-0.41\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-9}{9.85}=-0.91\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0)(2.77\cdot 2^{(x^{4})}x^{3}e^{(y^{2})}cos(z))+(-0.41)(2\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z))+(-0.91)(-1\cdot 2^{(x^{4})}e^{(y^{2})}sin(z))\&D_{\overrightarrow{u}}(x,y,z)=0.91\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) - 0.82\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z)\end{align*}$

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\&\text{Where }f(x,y,z)=4x^{4}cos(z^{3})ln(y^{2}) - 2\cdot2^{(4y)}\cdot4^xcos(4z)\end{align*}$

${32\cdot2^{(4y)}\cdot4^xcos(4z) - 24x^{4}zln(y^{2})sin(z^{3}) - 36x^{4}z^{4}cos(z^{3})ln(y^{2})}$

${-(8\cdot2^{(4y)}\cdot4^xsin(4z) - 12x^{4}z^{2}ln(y^{2})sin(z^{3}))\cdot(24x^{4}zln(y^{2})sin(z^{3}) - 32\cdot2^{(4y)}\cdot4^xcos(4z) + 36x^{4}z^{4}cos(z^{3})ln(y^{2}))}$

${16\cdot2^{(4y)}\cdot4^xsin(4z) - 24x^{4}z^{2}ln(y^{2})sin(z^{3})}$

${(8\cdot2^{(4y)}\cdot4^xsin(4z) - 12x^{4}z^{2}ln(y^{2})sin(z^{3}))^{2}}$

Explanation

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[sin(u)]=cos(u)du\&\text{Solve step-by-step:}\&f(x,y,z)=4x^{4}cos(z^{3})ln(y^{2}) - 2\cdot2^{(4y)}\cdot4^xcos(4z)\&f_{z}=8\cdot2^{(4y)}\cdot4^xsin(4z) - 12x^{4}z^{2}ln(y^{2})sin(z^{3})\&f_{zz}=32\cdot2^{(4y)}\cdot4^xcos(4z) - 24x^{4}zln(y^{2})sin(z^{3}) - 36x^{4}z^{4}cos(z^{3})ln(y^{2})\end{align*}$

Find the dot product between $\left \langle 3,-6,1\right \rangle$ and $\left \langle 1,4,5\right \rangle$

$\left \langle 3,-24,5\right \rangle$

Explanation

The formula for the dot product between two vectors $a=\left \langle x_1,y_1,z_1\right \rangle$ and $b=\left \langle x_2,y_2,z_2\right \rangle$ is $a\cdot b=(x_1*x_2)+(y_1*y_2)+(z_1*z_2)$ . Using the vectors in the problem statement, this becomes .

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (13\cdot (\frac{2}{7})^{(2x^{2} + 2y^{2})})}{6}-\frac{ (2\cdot 3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.48\text{ and }0.98\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.279,0.960)\text{ and }\overrightarrow{u_2}=(-0.982,-0.187)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(-\frac{ (13\cdot (\frac{2}{7})^{(2x^{2} + 2y^{2})})}{6}-\frac{ (2\cdot 3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})}\cdot r)}{6})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.960}{0.279})=0.41\pi;\theta_2=arctan(\frac{-0.187}{-0.982})=1.06\pi\&\text{Now, utilizing integral rules:}\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int_{0.41\pi}^{1.06\pi}\int_{0.48}^{0.98}(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})}\cdot r)}{6})drd\theta=(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})})}{(9ln(3))}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})})}{(24ln(\frac{2}{7}))})d\theta|_{0.48}^{0.98}\&\int_{0.41\pi}^{1.06\pi}(-0.8927)d\theta=(-0.8927\theta)|_{0.41\pi}^{1.06\pi}=-1.82\end{align*}$

Calculate antiderivative of $f(x)=\frac{3x^2+2x+1}{x^3+x^2+x+1}$

$\ln(x^3+x^2+x+1)+C$

$\ln(x^3+x^2+x)+C$

$\ln(x^3+x^2+x+20)+C$

$\ln(x^3+x^2+1)+C$

Explanation

We can calculate the antiderivative $\small f(x)=\frac{3x^2+2x+1}{x^3+x^2+x+1}$

$\small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx$

by using $\small u$ -substitution. We set $\small u=x^3+x^2+x+1$ , so we get $\small du=3x^2+2x+1$ , so the integral becomes

$\small \small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx=\int \frac{du}{u}=\ln u$

So we just plug $\small u=x^3+x^2+x+1$ to get

$\small \small \small \small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx=\int \frac{du}{u}=\ln u=\ln (x^3+x^2+x+1)$

$\begin{align*}&\text{Evaluate the triple integral}\int_{-2}^{3}\int_{7}^{11}\int_{-10}^{-6}(\frac{(43cos(x + 1)cos(4z)sin(y + 2))}{2})dxdydz\end{align*}$

Explanation

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\&\text{not entirely matter.}\&\text{For example:}\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\&\text{Considering our integral:}\&\int_{-2}^{3}\int_{7}^{11}\int_{-10}^{-6}(\frac{(43cos(x + 1)cos(4z)sin(y + 2))}{2})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{-2}^{3}\int_{7}^{11}\int_{-10}^{-6}(\frac{(43cos(x + 1)cos(4z)sin(y + 2))}{2})dxdydz=\int_{-2}^{3}\int_{7}^{11}(21.5cos(4z)sin(x + 1)sin(y + 2))dydz|_{-10}^{-6}\&\int_{-2}^{3}\int_{7}^{11}(29.48cos(4z)sin(y + 2))dydz=\int_{-2}^{3}(-29.48cos(y + 2)cos(4z))dz|_{7}^{11}\&\int_{-2}^{3}(-53.61cos(4z))dz=-13.4sin(4z)|_{-2}^{3}=-6.07\end{align*}$

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