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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #91 : Cross Product

Find the cross product of the vectors $\left \langle 10,-2,1\right \rangle$ and $\left \langle 5,-9,0\right \rangle$

Possible Answers:

$\left \langle 18,10,-81\right \rangle$

$\left \langle 9,5,-80\right \rangle$

$\left \langle 9,-5,-80\right \rangle$

$\left \langle 9,8,-40\right \rangle$

Correct answer:

$\left \langle 9,5,-80\right \rangle$

Explanation:

To find the cross product of two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you find the determinant of the 3x3 matrix $\begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1&a_2 &a_3 \\ b_1&b_2 & b_3 \end{bmatrix}$

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(0+9)-\hat{j}(0-5)+\hat{k}(-90+10)=\left \langle 9,5,-80\right \rangle$

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Example Question #92 : Cross Product

Find the cross product of the vectors $\left \langle 0,-1,3\right \rangle$ and $\left \langle 7,7,-2\right \rangle$

Possible Answers:

$\left \langle -19,21,7\right \rangle$

$\left \langle -19,20,1\right \rangle$

$\left \langle -16,2,7\right \rangle$

$\left \langle -19,-21,7\right \rangle$

Correct answer:

$\left \langle -19,21,7\right \rangle$

Explanation:

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(2-21)-\hat{j}(0-21)+\hat{k}(0+7)=\left \langle -19,21,7\right \rangle$

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Example Question #93 : Cross Product

Find the cross product of the vectors $\left \langle 9,-7,2\right \rangle$ and $\left \langle 3,-9,0\right \rangle$

Possible Answers:

$\left \langle 18,-6,-60\right \rangle$

$\left \langle 18,6,-10\right \rangle$

$\left \langle 8,16,-20\right \rangle$

$\left \langle 18,6,-60\right \rangle$

Correct answer:

$\left \langle 18,6,-60\right \rangle$

Explanation:

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(0+18)-\hat{j}(0-6)+\hat{k}(-81+21)=\left \langle 18,6,-60\right \rangle$

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Example Question #94 : Cross Product

Find the cross product of the vectors $\left \langle 10,3,13\right \rangle$ and $\left \langle 0,5,6\right \rangle$

Possible Answers:

$\left \langle -47,-60,50\right \rangle$

$\left \langle -47,-60,53\right \rangle$

$\left \langle -45,-65,55\right \rangle$

$\left \langle -47,-63,50\right \rangle$

Correct answer:

$\left \langle -47,-60,50\right \rangle$

Explanation:

To determine the cross product of two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we find the determinant of the 3x3 matrix $m=\begin{bmatrix} \hat{i}& \hat{j} & \hat{k}\\ a_1&a_2 &a_3 \\ b_1& b_2& b_3 \end{bmatrix}$ , using the formula

$determinant(m)=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using the vectors from the problem statement, we get

$\hat{i}(18-65)-\hat{j}(60-0)+\hat{k}(50-0)=\left \langle -47,-60,50\right \rangle$

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Example Question #95 : Cross Product

Find the cross product $\mathbf a\times \mathbf b$ , where

$\mathbf a = \left \langle 2,4,0\right \rangle$ and $\mathbf b \left \langle 5,-1,2\right \rangle$ .

Possible Answers:

$9\mathbf i-4 \mathbf j-18 \mathbf k$

$8\mathbf i-4 \mathbf j+22 \mathbf k$

$10\mathbf i+4 \mathbf j+18 \mathbf k$

$8\mathbf i+4 \mathbf j-18 \mathbf k$

$8\mathbf i-4 \mathbf j-22 \mathbf k$

Correct answer:

$8\mathbf i-4 \mathbf j-22 \mathbf k$

Explanation:

Recall the definition of the cross product $\mathbf a\times \mathbf b$ of two vectors $\mathbf a = \left \langle a_1, a_2, a_3\right \rangle$ and $\mathbf b = \left \langle b_1, b_2, b_3\right \rangle$ , in terms of determinants:

$\mathbf a\times \mathbf b= \begin{vmatrix} \mathbf i~\ \mathbf j~\ \mathbf k\\a_1\ a_2\ a_3\\b_1\ b_2\ b_3 \end{vmatrix}$

where $\mathbf i$ , $\mathbf j$ , and $\mathbf k$ are the standard basis vectors pointing in the directions of the positive -, - and -axes, respectively. We can apply this definition to calculate the cross product of $\mathbf a = \left \langle 2,4,0\right \rangle$ and $\mathbf b \left \langle 5,-1,2\right \rangle$ , as follows:

$\mathbf a\times \mathbf b= \begin{vmatrix} \mathbf i~\ \mathbf j~\ \mathbf k\\2~\ 4\ ~0\\5 -1\ 2 \end{vmatrix}$

$= \begin{vmatrix} 4\ \0\\ -1 \2\end{vmatrix} \mathbf i + \begin{vmatrix} 2\ ~0\\ 5 ~\2\end{vmatrix} \mathbf j + \begin{vmatrix} 2\ ~4\\ 5 -1\end{vmatrix} \mathbf k$

$=(8-0) \mathbf i-(4-0) \mathbf j+(-2-20) \mathbf k$

$=8 \mathbf i -4 \mathbf j-22 \mathbf k$

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Example Question #95 : Cross Product

Let u=(2,4,5) , and v=(-2,6,10) .

Find $u\times v$ .

Possible Answers:

$u\times v=(0,-3,2)$

$u\times v=(10,30,20)$

$u\times v=(10,-30,20)$

$u\times v=(40,-20,12)$

$u\times v=(-10,30,-20)$

Correct answer:

$u\times v=(10,-30,20)$

Explanation:

We are trying to find the cross product between and .

Recall the formula for cross product.

If u=(u_x,u_y,u_z) , and v=(v_x,v_y,v_z) , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

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Example Question #201 : Vectors And Vector Operations

Let u=(2,4,5) , and v=(-2,6,10) .

Find $u\times v$ .

Possible Answers:

$u\times v=(40,-20,12)$

$u\times v=(-10,30,-20)$

$u\times v=(10,-30,20)$

$u\times v=(10,30,20)$

$u\times v=(0,-3,2)$

Correct answer:

$u\times v=(10,-30,20)$

Explanation:

We are trying to find the cross product between and .

Recall the formula for cross product.

If u=(u_x,u_y,u_z) , and v=(v_x,v_y,v_z) , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

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Example Question #1 : Multiple Integration

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane z=x+1 , above the plane and between the cylinders x^2+y^2=1 , and x^2+y^2=9 .

Possible Answers:

$\frac{\pi}{4}$

$2 \pi$

$\pi$

$\frac{\pi}{2}$

Correct answer:

Explanation:

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above z=0 .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles x^2+y^2=1 , and x^2+y^2=9 .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r^2\sin(\theta) \ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} zr^2\sin(\theta) \Big|_{0}^{r\cos(\theta)+1} \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} (r\cos(\theta)+1)r^2\sin(\theta)-0 \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} r^3\sin(\theta)\cos(\theta)+r^2\sin(\theta) \ dr \ d\theta$

$=\int_{0}^{2\pi} \frac{1}{4}r^4\sin(\theta)\cos(\theta)+\frac{1}{3}r^3\sin(\theta) \Big|_{1}^{3}\ d\theta$

$=\int_{0}^{2\pi} (\frac{1}{4}(3)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(3)^3\sin(\theta))-(\frac{1}{4}(1)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(1)^3\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} (\frac{81}{4}\sin(\theta)\cos(\theta)+9\sin(\theta))-(\frac{1}{4}\sin(\theta)\cos(\theta)+\frac{1}{3}\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} 20\sin(\theta)\cos(\theta)+\frac{26}{3}\sin(\theta) \ d\theta$

$= -\frac{1}{2}\cdot20\cos^2(\theta)-\frac{26}{3}\cos(\theta)\Big|_{0}^{2\pi}$

$= -\frac{1}{2}\cdot20\cos^2(2\pi)-\frac{26}{3}\cos(2\pi)-(-\frac{1}{2}\cdot20\cos^2(0)-\frac{26}{3}\cos(0))$

$= -\frac{1}{2}\cdot20(1)-\frac{26}{3}(1)-(-\frac{1}{2}\cdot20(1)-\frac{26}{3}(1))$

$= -10-\frac{26}{3}-(-10-\frac{26}{3})=0$

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Example Question #2 : Triple Integrals

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

Possible Answers:

0.03

0.01

-0.07

-0.02

Correct answer:

0.03

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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Example Question #1 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}-\frac{ (49e^{(20z)}e^{(x^{2} + y^{2})})}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.92\\&\text{and length }1.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

$-2.08871\cdot10^{15}$

$1.39247\cdot10^{15}$

$-8.35483\cdot10^{15}$

$8.35483\cdot10^{15}$

Correct answer:

$-8.35483\cdot10^{15}$

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}-\frac{ (49e^{(20z)}e^{(x^{2} + y^{2})})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ (49\cdot re^{(r^{2})}e^{(20z)})}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.65}\int_{0.43\pi}^{1.73\pi}\int_{0}^{1.92}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ (49\cdot re^{(r^{2})}e^{(20z)})}{5})drd\theta dz=(-\frac{ (49e^{(20z + r^{2})})}{10}-\frac{ (3e^{(-\frac{(2\cdot r^{2})}{3})})}{16})d\theta dz|_{0}^{1.92}\\&\int_{0}^{1.65}\int_{0.43\pi}^{1.73\pi}(0.1714 - 190.6e^{(20z)})d\theta dz=(-1.0\theta\cdot (190.6e^{(20z)} - 0.1714))dz|_{0.43\pi}^{1.73\pi}\\&\int_{0}^{1.65}(0.7002 - 778.5e^{(20z)})dz=(0.7002z - 38.92e^{(20z)})|_{0}^{1.65}=-8.35483\cdot10^{15}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #91 : Cross Product

Example Question #92 : Cross Product

Example Question #93 : Cross Product

Example Question #94 : Cross Product

Example Question #95 : Cross Product

Example Question #95 : Cross Product

Example Question #201 : Vectors And Vector Operations

Example Question #1 : Multiple Integration

Example Question #2 : Triple Integrals

Example Question #1 : Triple Integration In Cylindrical Coordinates

All Calculus 3 Resources

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