We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

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

Example Questions

Example Question #201 : Vectors And Vector Operations

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

Possible Answers:

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

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

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

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

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:

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

Report an Error

Example Question #93 : 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:

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

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

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

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

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:

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

Report an Error

Example Question #94 : 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:

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

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

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

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

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:

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

Report an Error

Example Question #95 : 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:

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

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

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

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

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

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

Report an Error

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:

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

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

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

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

Report an Error

Example Question #451 : Calculus 3

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

Find $u\times v$ .

Possible Answers:

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

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

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

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

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

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.

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

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

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

Report an Error

Example Question #451 : Calculus 3

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

Find $u\times v$ .

Possible Answers:

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

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

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

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

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

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.

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

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

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

Report an Error

Example Question #1 : Triple Integrals

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:

$\frac{\pi}{2}$

$2 \pi$

$\frac{\pi}{4}$

$\pi$

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

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

$1\leq r \leq 3$

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

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

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

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

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

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

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

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

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

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

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

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

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

Report an Error

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:

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

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

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

Report an Error

Example Question #2 : Triple Integrals

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

Possible Answers:

$1.39247\cdot10^{15}$

$8.35483\cdot10^{15}$

$-8.35483\cdot10^{15}$

$-2.08871\cdot10^{15}$

Correct answer:

$-8.35483\cdot10^{15}$

Explanation:

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

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

Report an Error

View Calculus 3 Tutors

Fatemeh
Certified Tutor

University of Isfahan, Doctor of Engineering, Chemical Engineering. University of Windsor, Doctor of Engineering, Environment...

View Calculus 3 Tutors

Michael
Certified Tutor

University of California-San Diego, Bachelors, Physics with specialization in Earth Science. University of California-San Die...

View Calculus 3 Tutors

Madeline
Certified Tutor

University of Central Florida, Current Undergrad, Physics.

All Calculus 3 Resources

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

Popular Subjects

GRE Tutors in Miami, Chemistry Tutors in Phoenix, Physics Tutors in Dallas Fort Worth, Computer Science Tutors in Boston, Biology Tutors in Atlanta, MCAT Tutors in Denver, GMAT Tutors in Chicago, Math Tutors in New York City, ACT Tutors in Houston, LSAT Tutors in Boston

Popular Courses & Classes

GRE Courses & Classes in Seattle, GRE Courses & Classes in Atlanta, LSAT Courses & Classes in Atlanta, GMAT Courses & Classes in Atlanta, GMAT Courses & Classes in Boston, SAT Courses & Classes in Los Angeles, SAT Courses & Classes in Chicago, LSAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Chicago, GRE Courses & Classes in Phoenix

Popular Test Prep

GMAT Test Prep in Atlanta, SAT Test Prep in New York City, GMAT Test Prep in Chicago, LSAT Test Prep in Philadelphia, GMAT Test Prep in Washington DC, GRE Test Prep in Seattle, SAT Test Prep in Philadelphia, LSAT Test Prep in Miami, LSAT Test Prep in Dallas Fort Worth, ACT Test Prep in Phoenix

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #201 : Vectors And Vector Operations

Example Question #93 : Cross Product

Example Question #94 : Cross Product

Example Question #95 : Cross Product

Example Question #95 : Cross Product

Example Question #451 : Calculus 3

Example Question #451 : Calculus 3

Example Question #1 : Triple Integrals

Example Question #2 : Triple Integrals

Example Question #2 : Triple Integrals

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: