You CAN Ace Calculus

 derivatives integrals basics of polar coordinates

### 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.

On this page we cover the most common calculus problems involving polar coordinates. The two main topics are differentiation and integration.
As mentioned on the main polar coordinates page, polar coordinates are just parametric equations. If you are familiar with parametric equations, this material should be very intuitive.

Differentiation - Slope and Tangent Lines

In order to find slope and, subsequently, a tangent line to a smooth graph, we need to be able to calculate the derivative of a parametric equation.

We have the graph represented by the parametric equations $$x = r(\theta)\cos(\theta)$$ and $$y = r(\theta)\sin(\theta)$$. Here we are emphasizing that r is a a function of $$\theta$$ by writing $$r(\theta)$$. The derivative is given by

$$\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{r' \sin(\theta) + r \cos(\theta)}{r' \cos(\theta) - r \sin(\theta)} }$$     where $$r' = dr/d\theta$$.

### Polar Slope Proof

Theorem: Polar Slope

For a set of polar equations $$x = r \cos(\theta)$$ and $$y = r \sin(\theta)$$ where $$r$$ is a function of $$\theta$$, the slope $$dy/dx$$ of a smooth function is given by

$$\displaystyle{\frac{dy}{dx} = \frac{r'\sin(\theta) + r \cos(\theta)}{r'\cos(\theta) - r \sin(\theta)}}$$

where $$r' = dr/d\theta$$

To prove this, we use the theorem of parametrics derivative, the result of which is $$\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }$$

In our case, the parameter is $$\theta$$ and our parametric equations are $$x(\theta) = r(\theta)\cos(\theta)$$ and $$y(\theta) = r(\theta)\sin(\theta)$$. So our parametric derivative is $$\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} }$$

Using the chain rule, we calculate
$$\displaystyle{\frac{dy}{d\theta} = r'\sin(\theta) + r \cos(\theta)}$$ and $$\displaystyle{ \frac{dx}{d\theta} = r'\cos(\theta) - r \sin(\theta)}$$

Putting these results in the parametric derivative equation, we have

$$\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{r'\sin(\theta) + r \cos(\theta)}{r'\cos(\theta) - r \sin(\theta)} }$$ [qed]

This equation gives us the slope of the graph at every point. Once we have the slope, we can calculate the equation of a tangent line using the technique found on the main tangent page. Before we go on, let's work some practice problems.

Basic Problems

Calculate the slope of the tangent line to the curve $$r=4$$ at $$\theta=\pi/4$$.

Problem Statement

Calculate the slope of the tangent line to the curve $$r=4$$ at $$\theta=\pi/4$$.

Solution

### 1269 solution video

video by PatrickJMT

Calculate the slope of the tangent line to the curve $$r=1+5\cos\theta$$ at $$\theta=\pi/4$$.

Problem Statement

Calculate the slope of the tangent line to the curve $$r=1+5\cos\theta$$ at $$\theta=\pi/4$$.

Solution

### 1270 solution video

video by PatrickJMT

Calculate the slope of the tangent line to the curve $$r=\sin(5\theta)$$ at $$\theta=\pi/5$$.

Problem Statement

Calculate the slope of the tangent line to the curve $$r=\sin(5\theta)$$ at $$\theta=\pi/5$$.

Solution

### 1271 solution video

video by PatrickJMT

Find the tangent line to the curve $$r=\cos(2\theta)$$ at $$\theta=\pi/4$$.

Problem Statement

Find the tangent line to the curve $$r=\cos(2\theta)$$ at $$\theta=\pi/4$$.

Solution

### 1279 solution video

video by Krista King Math

Intermediate Problems

Find the points on the polar curve $$r=3\cos(\theta)$$ where the graph of the tangent line is vertical or horizontal.

Problem Statement

Find the points on the polar curve $$r=3\cos(\theta)$$ where the graph of the tangent line is vertical or horizontal.

Solution

### 1280 solution video

video by Krista King Math

Integration - Area

To calculate the area defined by a polar curve r, which is a function of $$\theta$$, between the rays $$\theta = \theta_0$$ and $$\theta = \theta_1$$, we use the integral

$$\displaystyle{A = \frac{1}{2} \int_{\theta_0}^{\theta_1}{ r^2 d\theta}}$$

Here is a good in-depth video discussing area using polar coordinates.

### MIT OCW - Lec 33 | MIT 18.01 Single Variable Calculus, Fall 2007 [35min]

video by MIT OCW

Basic Problems

Calculate the area enclosed by the polar curve $$r=2\cos\theta$$.

Problem Statement

Calculate the area enclosed by the polar curve $$r=2\cos\theta$$.

Solution

### 1383 solution video

video by Krista King Math

Calculate the area bounded by one loop of $$r=2\cos(4\theta)$$.

Problem Statement

Calculate the area bounded by one loop of $$r=2\cos(4\theta)$$.

Solution

### 1384 solution video

video by Krista King Math

Calculate the area enclosed by one loop of $$r=\sin(4\theta)$$.

Problem Statement

Calculate the area enclosed by one loop of $$r=\sin(4\theta)$$.

Solution

### 1272 solution video

video by PatrickJMT

Calculate the area enclosed by $$r=3+\sin\theta$$.

Problem Statement

Calculate the area enclosed by $$r=3+\sin\theta$$.

Solution

### 1276 solution video

video by PatrickJMT

Intermediate Problems

Calculate the area inside $$r=1+2\cos(\theta)$$ and outside $$r=2$$.

Problem Statement

Calculate the area inside $$r=1+2\cos(\theta)$$ and outside $$r=2$$.

Solution

### 1382 solution video

video by Krista King Math

Calculate the area inside both $$r=\cos\theta$$ and $$r=\sin(2\theta)$$.

Problem Statement

Calculate the area inside both $$r=\cos\theta$$ and $$r=\sin(2\theta)$$.

$$\displaystyle{ \frac{\pi}{4} - \frac{3\sqrt{3}}{16} }$$

Problem Statement

Calculate the area inside both $$r=\cos\theta$$ and $$r=\sin(2\theta)$$.

Solution

The final answer at the end of the video is incorrect. At about the 11:26 mark he notes his mistake.

### 1273 solution video

video by PatrickJMT

$$\displaystyle{ \frac{\pi}{4} - \frac{3\sqrt{3}}{16} }$$

Calculate the region that lies inside $$r=1+\sin\theta$$ and outside $$r=1$$.

Problem Statement

Calculate the region that lies inside $$r=1+\sin\theta$$ and outside $$r=1$$.

Solution

### 1274 solution video

video by PatrickJMT

Calculate the area of the region inside both $$r=9\cos\theta$$ and $$r=9\sin\theta$$.

Problem Statement

Calculate the area of the region inside both $$r=9\cos\theta$$ and $$r=9\sin\theta$$.

Solution

### 1275 solution video

video by PatrickJMT

Calculate the area outside $$r=\sin\theta$$ and inside $$r=2\sin\theta$$.

Problem Statement

Calculate the area outside $$r=\sin\theta$$ and inside $$r=2\sin\theta$$.

Solution

### 1277 solution video

video by PatrickJMT

Calculate the area of the region inside both $$r=\sin(2\theta)$$ and $$r=\cos(2\theta)$$.

Problem Statement

Calculate the area of the region inside both $$r=\sin(2\theta)$$ and $$r=\cos(2\theta)$$.

Solution

### 1278 solution video

video by Krista King Math

Integration - Arc Length

To calculate the arc length of a smooth curve, we can use this integral.

$$\displaystyle{s = \int_{\theta_0}^{\theta_1}{\sqrt{r^2 + [r']^2} ~ d\theta}}$$
where r is a function of $$\theta$$, $$r' = dr/d\theta$$ and we are looking at the arc from $$\theta = \theta_0$$ to $$\theta = \theta_1$$.

Basic Problems

Calculate the arc length of the polar curve $$r=e^{\theta/2}$$, $$0\leq\theta\leq4\pi$$.

Problem Statement

Calculate the arc length of the polar curve $$r=e^{\theta/2}$$, $$0\leq\theta\leq4\pi$$.

Solution

### 1385 solution video

video by Krista King Math

Intermediate Problems

Calculate the arc length of the polar curve $$r=\sin^2(\theta/2)$$, $$0\leq\theta\leq\pi$$.

Problem Statement

Calculate the arc length of the polar curve $$r=\sin^2(\theta/2)$$, $$0\leq\theta\leq\pi$$.

Solution

### 1386 solution video

video by Krista King Math

Integration - Surface Area

To calculate the surface area of a polar curve revolved about an axis, we use these integrals.

The equation of the polar curve is in the form $$x=r(\theta) \cos(\theta)$$ and $$y = r(\theta) \sin(\theta)$$ and the curve that is being revolved is defined to be from $$\theta_0$$ to $$\theta_1$$. We also define $$ds = \sqrt{r^2 + [dr/d\theta]^2}~dt$$.

surface area

(the polar axis)

$$\displaystyle{ S = 2\pi \int_{\theta_0}^{\theta_1}{y~ds} = }$$ $$\displaystyle{ 2\pi \int_{\theta_0}^{\theta_1}{ r(\theta) \sin(\theta) \sqrt{r^2 + [dr/d\theta]^2}~dt } }$$

$$\displaystyle{ S = 2\pi \int_{\theta_0}^{\theta_1}{x~ds} = }$$ $$\displaystyle{ 2\pi \int_{\theta_0}^{\theta_1}{ r(\theta) \cos(\theta) \sqrt{r^2 + [dr/d\theta]^2}~dt } }$$

So, why the $$ds$$ term? If you look at the previous section on arc length, you will notice that this term appears in that integral. This is called a differential length and is just a convenient way of writing these integrals.

Intermediate Problems

Calculate the surface area formed by rotating the polar curve $$r=4\sin\theta$$, $$0\leq\theta\leq\pi$$ about the x-axis.

Problem Statement

Calculate the surface area formed by rotating the polar curve $$r=4\sin\theta$$, $$0\leq\theta\leq\pi$$ about the x-axis.

Solution

### 1268 solution video

video by Krista King Math