You CAN Ace Calculus

### Topics You Need To Understand For This Page

 integration by parts trig integration integration using trig substitution partial fractions volume of revolution centroids work parametrics

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

This is the first exam for second semester single variable calculus. The exam covers advanced integration and applications including work, volume of revolution and centroids. Also parametrics are part of this exam.

### Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

1.5 hours

Questions

10

Total Points

100

Tools

Calculator

see instructions

Formula Sheet(s)

none

Other Tools

none

Instructions:
- This exam is in two main parts, labeled parts A and B, with different instructions for each part.
- Show all your work.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).
- Give exact, simplified answers.

### Part A - Questions 1-4

Instructions for Part A - - You have 30 minutes to complete this part of the exam. No calculators are allowed. Show all your work, justify your conclusions and give exact, completely factored answers.

Section 1

Evaluate the following integrals. Each question in this section is worth 9 points.

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } }$$

Final Answer

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } = \frac{e^4-13}{32e^4} }$$

Problem Statement

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } }$$

Solution

Use integration by parts: $$\int{u~dv} = uv - \int{ v ~du }$$

 $$u=x^2 \to du = 2x ~dx$$ $$dv=e^{-4x}dx \to \displaystyle{v=\frac{e^{-4x}}{-4}}$$

$$\displaystyle{\left. \frac{-x^2}{4}e^{-4x} \right|_0^1 - \int_{0}^{1}{\left(\frac{-1}{4} e^{-4x} \right) 2x~dx} =}$$ $$\displaystyle{\frac{-1}{4}e^{-4} + 0 + \frac{1}{2} \int_{0}^{1}{xe^{-4x}dx}}$$
Use integration by parts again.

 $$\begin{array}{lcl} u=x & \to & du=dx \\ dv = e^{-4x}dx & \to & \displaystyle{ v=\frac{e^{-4x}}{-4} } \end{array}$$

$$\displaystyle{\frac{-1}{4e^4} + \frac{1}{2}\left[ \left. \frac{-x}{4}e^{-4x} \right|_0^1 - \int_{0}^{1}{\frac{-1}{4}e^{-4x}dx} \right]}$$
$$\displaystyle{\frac{-1}{4e^4} + \frac{1}{2}\left[ \frac{-1}{4}e^{-4} +0 \right] + \frac{1}{8} \int_{0}^{1}{e^{-4x}dx}}$$
$$\displaystyle{\frac{-3}{8e^4} + \left. \frac{1}{8}\frac{e^{-4x}}{4} \right|_0^1}$$
$$\displaystyle{\frac{-3}{8e^4} - \frac{1}{32}\left[ e^{-4} - 1 \right] = \frac{e^4-13}{32e^4}}$$

Final Answer

$$\displaystyle{ \int_{0}^{1}{ x^2e^{-4x} ~dx } = \frac{e^4-13}{32e^4} }$$

$$\int{ \sin^3(\pi t) \cos^3(\pi t) ~dt }$$

Problem Statement

$$\int{ \sin^3(\pi t) \cos^3(\pi t) ~dt }$$

Final Answer

$$\displaystyle{ \int{ \sin^3(\pi t)\cos^3(\pi t) ~dt } = }$$ $$\displaystyle{ \frac{\cos^4(\pi t )}{12}(2\cos^2(\pi t)-3) + C }$$

Problem Statement

$$\int{ \sin^3(\pi t) \cos^3(\pi t) ~dt }$$

Solution

$$\int{\sin(\pi t)\sin^2(\pi t)\cos^3(\pi t)~dt}$$
$$\int{\sin(\pi t)(1-\cos^2(\pi t))\cos^3(\pi t)~dt}$$

 $$u=\cos(\pi t)$$ $$du=-\sin(\pi t)dt$$ $$-\int{(1-u^2)u^3du}$$ $$\int{u^5-u^3 du}$$ $$\displaystyle{\frac{u^6}{6}-\frac{u^4}{4}+C}$$ $$\displaystyle{\frac{u^4}{12}(2u^2-3)+C}$$ $$\displaystyle{\frac{\cos^4(\pi t)}{12}(2\cos^2(\pi t)-3)+C}$$

Final Answer

$$\displaystyle{ \int{ \sin^3(\pi t)\cos^3(\pi t) ~dt } = }$$ $$\displaystyle{ \frac{\cos^4(\pi t )}{12}(2\cos^2(\pi t)-3) + C }$$

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } }$$

Final Answer

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } = \frac{3\sqrt{2}}{8} }$$

Problem Statement

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } }$$

Solution

Use $$1+\tan^2(\theta)=\sec^2(\theta)$$ and let $$r=2\tan(\theta) ~~ \to ~~ dr=2\sec^2(\theta)~d\theta$$.
The triangle based on $$r=2\tan(\theta)$$ is shown to the right. Using the Pythagorean Theorem, the hypotenuse is $$\displaystyle{\sqrt{r^2+4}}$$.
First, we will evaluate the integral without the limits of integration (the indefinite integral).

 $$\displaystyle{\int{\frac{3}{(r^2+4)^{3/2}}dr}}$$ $$\displaystyle{\int{\frac{3}{(4\tan^2(\theta) + 4)^{3/2}}2\sec^2(\theta)d\theta}}$$ $$\displaystyle{\int{\frac{6\sec^2(\theta)}{4^{3/2}(1+\tan^2(\theta))^{3/2}}d\theta}}$$ $$\displaystyle{\frac{3}{4}\int{\frac{\sec^2(\theta)}{(\sec^2(\theta))^{3/2}}d\theta}}$$ $$\displaystyle{\frac{3}{4}\int{\frac{1}{\sec(\theta)}d\theta}}$$ $$\displaystyle{\frac{3}{4}\int{\cos(\theta)~d\theta}}$$ $$\displaystyle{\frac{3}{4}\sin(\theta)+C}$$ Use the triangle to convert from $$\theta$$ to $$r$$. $$\displaystyle{\frac{3}{4}\frac{r}{\sqrt{r^2+4}}+C}$$

Now we will evaluate the integral using the limits of integration.
$$\displaystyle{\left. \frac{3}{4} \frac{r}{\sqrt{r^2+4}} \right|_0^2 = \frac{3}{4} \frac{2}{\sqrt{8}} - 0 = \frac{3}{2} \frac{1}{2\sqrt{2}} = \frac{3\sqrt{2}}{8} }$$

Final Answer

$$\displaystyle{ \int_{0}^{2}{ \frac{3}{(r^2+4)^{3/2}} dr } = \frac{3\sqrt{2}}{8} }$$

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)} dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)} dx } }$$

Final Answer

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)}dx} = }$$ $$\displaystyle{ \ln\abs{(x+1)^3(x^2+4)^3} + k }$$

Problem Statement

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)} dx } }$$

Solution

Use partial fraction expansion.
$$\displaystyle{\frac{9x^2+6x+12}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+4}}$$
$$9x^2+6x+12=A(x^2+4)+(Bx+C)(x+1)$$
$$x=-1$$: $$~~ 9-6+12 = A(1+4)$$ $$\to$$ $$15=A(5)$$ $$\to$$ $$A=3$$
$$x=0$$: $$~~ 12=A(4)+C(1)$$ $$\to$$ $$12=12+C$$ $$\to$$ $$C=0$$
$$x=1$$: $$~~ 9+6+12=A(5)+B(1+1)$$ $$\to$$ $$27=15+2B$$ $$\to$$ $$(27-15)/2=B$$ $$\to$$ $$B=6$$
The integral can now be written
$$\displaystyle{\int{\frac{3}{x+1}+\frac{6x}{x^2+4}dx}}$$
The first fraction is easily integrated.
$$\displaystyle{\int{\frac{3}{x+1}dx}=3\ln\abs{x+1}+c_1}$$
Use substitution for the second fraction.
$$u=x^2+4 ~~ \to ~~ du=2x~dx$$
$$\displaystyle{\int{\frac{6x}{x^2+4}dx}=}$$ $$\displaystyle{\int{\frac{3~du}{u}dx}=}$$ $$\displaystyle{3\ln\abs{u}+c_2=}$$ $$\displaystyle{3\ln\abs{x^2+4}+c_2}$$
Putting the results together, we get
$$3\ln\abs{x+1}+3\ln\abs{x^2+4}+k=$$ $$3\ln\abs{(x+1)(x^2+4)}+k=$$ $$\ln\abs{(x+1)^3(x^2+4)^3}+k$$
We combined the constants $$k=c_1+c_2$$.

Final Answer

$$\displaystyle{ \int{ \frac{9x^2+6x+12}{(x+1)(x^2+4)}dx} = }$$ $$\displaystyle{ \ln\abs{(x+1)^3(x^2+4)^3} + k }$$

### Part B - Questions 5-10

Instructions for Part B - - You may use your calculator. You have one hour to complete this part of the exam. Show all your work. Correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions. Give exact, simplified answers.

Section 2 [16 points]

A region $$R$$ is bounded by the graph of $$y = e^x$$, the line $$y=1$$ and the line $$x=3$$.
Sketch the region $$R$$. [ click here to show/hide the plot ]

A solid is generated by revolving the region $$R$$ about the line $$x=-1$$. Set up, but do not evaluate, a definite integral for the volume of the resulting solid of revolution.

Problem Statement

A solid is generated by revolving the region $$R$$ about the line $$x=-1$$. Set up, but do not evaluate, a definite integral for the volume of the resulting solid of revolution.

Solution

Since the problem statement did not specify which method, we can use either the disc or shell method. Both solutions are shown here.

Disc Method

$$\displaystyle{V=\pi\int_c^d{R^2-r^2~dy}}$$
$$R=4 ~~~~ r=x+1$$
$$y=e^x ~~ \to ~~ \ln(y)=x ~~ \to ~~ r=\ln(y)+1$$
$$\displaystyle{V=\pi\int_1^{e^3}{16-[1+\ln(y)]^2~dy}}$$

Shell Method

$$\displaystyle{V=2\pi\int_a^b{rh~dx}}$$
$$r=1+x ~~~~ h=y-1=e^x-1$$
$$\displaystyle{V=2\pi\int_0^3{(1+x)(e^x-1)~dx}}$$

A second solid has base $$R$$ and cross-sections perpendicular to the x-axis are disks with a diameter in $$R$$. Set up, but do not evaluate, a definite integral for the volume of this region.

Problem Statement

A second solid has base $$R$$ and cross-sections perpendicular to the x-axis are disks with a diameter in $$R$$. Set up, but do not evaluate, a definite integral for the volume of this region.

Solution

Disks mean that we have circles with diameter $$d$$, so the radius is $$r=d/2$$. The area of a disk is $$\pi r^2$$. The radius is determined by cross-sections perpendicular to the x-axis, which are dependent on $$x$$.
$$d=y-1=e^x-1 ~~ \to ~~ r=(e^x-1)/2$$
$$\displaystyle{V=\int_0^3{\pi \left( \frac{e^x-1}{2}\right)^2 dx}}$$

Section 3

Solve the following problems. Make sure your answers have correct units.

(16 points) A thin metal plate in the plane occupies the triangle with vertices $$(0,2)$$, $$(0,0)$$ and $$(2,0)$$. Find the centroid of the plate. You may use symmetry and known area or volume formulas from geometry to evaluate integrals.

Problem Statement

(16 points) A thin metal plate in the plane occupies the triangle with vertices $$(0,2)$$, $$(0,0)$$ and $$(2,0)$$. Find the centroid of the plate. You may use symmetry and known area or volume formulas from geometry to evaluate integrals.

Final Answer

$$\displaystyle{\left(\frac{2}{3},\frac{2}{3}\right)}$$

Problem Statement

(16 points) A thin metal plate in the plane occupies the triangle with vertices $$(0,2)$$, $$(0,0)$$ and $$(2,0)$$. Find the centroid of the plate. You may use symmetry and known area or volume formulas from geometry to evaluate integrals.

Solution

From symmetry we can see that $$\bar{x}=\bar{y}$$.
The equation for centroid is
$$\displaystyle{\bar{x}=\frac{\text{moment about x}}{\text{area}}=\frac{\int_a^b{x(y_2-y_1)~dx}}{\text{area}}}$$
where $$y_1=0$$ and $$y_2=2-x$$
The area of the triangle is $$A=bh/2=(2)(2)/2=2$$

 $$\displaystyle{\int_0^2{x(2-x-0)~dx}}$$ $$\displaystyle{\int_0^2{2x-x^2~dx}}$$ $$\displaystyle{\left[ \frac{2x^2}{2}-\frac{x^3}{3}\right]_0^2}$$ $$\displaystyle{4-\frac{8}{3}=\frac{4}{3}}$$

$$\displaystyle{\bar{x}=\frac{4/3}{2}=\frac{2}{3}}$$

Final Answer

$$\displaystyle{\left(\frac{2}{3},\frac{2}{3}\right)}$$

(16 points) An upright water tank has a flat base in the plane and is filled with water. The tank has height $$7$$ feet, and, for $$0\le y\le 7$$, the cross-section parallel to the base at height $$y$$ has area $$y^2+2$$. How much work is done in pumping all the water over the top edge of the tank? [Use $$\delta~ lb/ft^3$$ for the weight-density of water.]

Problem Statement

(16 points) An upright water tank has a flat base in the plane and is filled with water. The tank has height $$7$$ feet, and, for $$0\le y\le 7$$, the cross-section parallel to the base at height $$y$$ has area $$y^2+2$$. How much work is done in pumping all the water over the top edge of the tank? [Use $$\delta~ lb/ft^3$$ for the weight-density of water.]

Final Answer

$$\displaystyle{ W = \frac{2989\delta}{12} }$$ ft-lbs

Problem Statement

(16 points) An upright water tank has a flat base in the plane and is filled with water. The tank has height $$7$$ feet, and, for $$0\le y\le 7$$, the cross-section parallel to the base at height $$y$$ has area $$y^2+2$$. How much work is done in pumping all the water over the top edge of the tank? [Use $$\delta~ lb/ft^3$$ for the weight-density of water.]

Solution

The picture to the right is meant to represent an upright tank filled with water. The $$dy$$ in the picture is a thin differential slice of water with cross sectional area is given in the problem statement as $$A=y^2+2$$.
To calculate the work, we know that the work, $$W$$ is force times distance, $$W=F\cdot d$$. To get force, we multiply the volume of one small slice of water times the weight-density.
The volume of one slice of water is $$dV=(y^2+2)dy ~ ft^3$$.
The force of one slice of water is $$dF=\delta\cdot dV=\delta(y^2+2)dy~lbs$$
Now, the distance $$d$$ is the distance one slice of the water is moved up over the top of the tank. The very top slice of water is moved a distance $$d=0~ft$$ while the slice at the bottom of the tank is moved $$d=7~ft$$. So our expression for the distance of any given slice of water is dependent on $$y$$ and is $$d=7-y$$. Now we are ready to set up our integral.
$$\displaystyle{\int_0^7{\delta (y^2+2)(7-y)dy}}$$
We integrate from the bottom of the tank ($$y=0)$$ to the top of the tank $$(y=7)$$ since the tank is filled to the top with water.

 $$\displaystyle{\int_0^7{\delta(y^2+2)(7-y)dy}}$$ $$\displaystyle{\int_0^7{7y^2-y^3+14-2y~dy}}$$ $$\displaystyle{\delta \left[ \frac{7y^3}{3}-\frac{y^4}{4}+14y-\frac{2y^2}{2}\right]_0^7}$$ $$\displaystyle{\delta \left[ \frac{7}{3}(7^3)-\frac{7^4}{4}+14(7)-7^2\right]-0}$$ $$\displaystyle{\frac{49(61)\delta}{12}}$$

Final Answer

$$\displaystyle{ W = \frac{2989\delta}{12} }$$ ft-lbs

Section 4

Let G be the portion of the graph with parametric equations
$$x= 1+\sin(t)$$     $$y = 3+2\cos(t)$$    for $$0 \le t \le 2\pi$$.
Plot G and answer the following questions based on the graph.
[ click here to show/hide the plot of G ]

(8 points) Set up, but do not evaluate, the integral for the length of G.

Problem Statement

(8 points) Set up, but do not evaluate, the integral for the length of G.

Final Answer

$$\displaystyle{ s = \int_0^{2\pi}{ \sqrt{\cos^2t+4\sin^2t} ~dt } }$$

Problem Statement

(8 points) Set up, but do not evaluate, the integral for the length of G.

Solution

The general equation is $$\displaystyle{s=\int_a^b{\sqrt{[x'(t)]^2+[y(t)]^2}~dt}}$$
In this question
$$x(t)=1+\sin(t) ~~ \to ~~ x'(t)=\cos(t)$$
$$y(t)=3+2\cos(t) ~~ \to ~~ y'(t)=-2\sin(t)$$
So our integral is $$\displaystyle{s=\int_0^{2\pi}{\sqrt{\cos^2t+4\sin^2t}~dt}}$$

Final Answer

$$\displaystyle{ s = \int_0^{2\pi}{ \sqrt{\cos^2t+4\sin^2t} ~dt } }$$

(8 points) Set up, but do not evaluate, the integral for the area of the surface generated by revolving G about the x-axis.

Problem Statement

(8 points) Set up, but do not evaluate, the integral for the area of the surface generated by revolving G about the x-axis.

Final Answer

$$\displaystyle{ A_x = 2\pi \int_0^{2\pi}{ (3+2\cos(t))\sqrt{\cos^2t+4\sin^2t} ~dt } }$$

Problem Statement

(8 points) Set up, but do not evaluate, the integral for the area of the surface generated by revolving G about the x-axis.

Solution

 The general equation is $$\displaystyle{ A_x = 2\pi \int_a^b{ y(t)\sqrt{(dx/dt)^2+(dy/dt)^2} ~dt } }$$

From the previous question, we know $$dx/dt$$ and $$dy/dt$$. So we just need to write down the integral.
$$\displaystyle{A_x=2\pi \int_0^{2\pi}{(3+2\cos(t))\sqrt{\cos^2t+4\sin^2t }~dt}}$$

Final Answer

$$\displaystyle{ A_x = 2\pi \int_0^{2\pi}{ (3+2\cos(t))\sqrt{\cos^2t+4\sin^2t} ~dt } }$$