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 polar coordinates double integrals

double integrals in polar coordinates youtube playlist

### 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

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17calculus > partial integrals > double integrals in polar coordinates

This page covers double integrals in polar coordinates. Double integrals in rectangular coordinates are covered on a separate page.

As you learned on the polar coordinates page, you use the equations $$x=r\cos\theta$$ and $$y=r\sin\theta$$ to convert equations from rectangular to polar coordinates. The same idea applies to a function and the description of an area in the xy-plane.

For example, if you have an integral in rectangular coordinates that looks like $$\iint_A{f(x,y)~dA}$$, you need to do three things to convert this to polar coordinates.
1. First, substitute for x and y in $$f(x,y)$$ using the above equations to get $$f(r\cos\theta,r\sin\theta)$$. This new form of the function can be written as $$f(r,\theta)$$.
2. Second, describe the area in polar coordinates or, if the area is already given in rectangular coordinates, convert the area in the xy-plane from rectangular to polar coordinates.
3. Finally, set up the integral with the function $$f(r,\theta)$$ in polar coordinates, being careful to integrate in the correct order.

Okay, so that is the big picture but how do you implement this when working problems? Time for some videos. Both of these videos are rather long but they will give you a good handle on double integrals in polar coordinates. You do not need to watch both of them since they cover the same ideas. But if you do, you will have a better understanding of the techniques.

video by MIT OCW

### Evans Lawrence - Multivariable Calculus: Lecture 19 - Double Integration in Polar Coordinates [35mins-2secs]

video by Evans Lawrence

### Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, evaluate the following integrals using polar coordinates, giving your answers in exact terms.

Basic Problems

$$\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1-y^2}}{ 3y~dx~dy} } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1-y^2}}{ 3y~dx~dy} } }$$

$$1 - 1/\sqrt{2}$$

Problem Statement

$$\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1-y^2}}{ 3y~dx~dy} } }$$

Solution

### 1950 solution video

video by Dr Chris Tisdell

$$1 - 1/\sqrt{2}$$

$$\displaystyle{ \int_{0}^{2}{ \int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }$$

Problem Statement

$$\displaystyle{ \int_{0}^{2}{ \int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }$$

$$32/9$$

Problem Statement

$$\displaystyle{ \int_{0}^{2}{ \int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }$$

Solution

### 1951 solution video

video by Dr Chris Tisdell

$$32/9$$

Determine the volume of the solid below the surface $$f(x,y) = 4 - x^2 - y^2$$ above the xy-plane over the region bounded by $$x^2 + y^2 = 1$$ and $$x^2 + y^2 = 4$$.

Problem Statement

Determine the volume of the solid below the surface $$f(x,y) = 4 - x^2 - y^2$$ above the xy-plane over the region bounded by $$x^2 + y^2 = 1$$ and $$x^2 + y^2 = 4$$.

$$9 \pi/2$$

Problem Statement

Determine the volume of the solid below the surface $$f(x,y) = 4 - x^2 - y^2$$ above the xy-plane over the region bounded by $$x^2 + y^2 = 1$$ and $$x^2 + y^2 = 4$$.

Solution

### 1952 solution video

video by MIP4U

$$9 \pi/2$$

Determine the volume of $$z = \sqrt{9-x^2-y^2}$$ over the region $$x^2 + y^2 \leq 4$$ in the first octant.

Problem Statement

Determine the volume of $$z = \sqrt{9-x^2-y^2}$$ over the region $$x^2 + y^2 \leq 4$$ in the first octant.

$$(27-5\sqrt{5}) \pi/6$$

Problem Statement

Determine the volume of $$z = \sqrt{9-x^2-y^2}$$ over the region $$x^2 + y^2 \leq 4$$ in the first octant.

Solution

### 1953 solution video

video by MIP4U

$$(27-5\sqrt{5}) \pi/6$$

Evaluate $$\iint_{A}{ e^{-x^2-y^2} ~dA }$$ where A is bounded by $$x = \sqrt{4-y^2}$$ and the y-axis.

Problem Statement

Evaluate $$\iint_{A}{ e^{-x^2-y^2} ~dA }$$ where A is bounded by $$x = \sqrt{4-y^2}$$ and the y-axis.

$$(\pi/2)(1-1/e^4)$$

Problem Statement

Evaluate $$\iint_{A}{ e^{-x^2-y^2} ~dA }$$ where A is bounded by $$x = \sqrt{4-y^2}$$ and the y-axis.

Solution

### 1954 solution video

video by Krista King Math

$$(\pi/2)(1-1/e^4)$$

$$\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }$$

Problem Statement

$$\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }$$

$$\sqrt{2}/4$$

Problem Statement

$$\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }$$

Solution

### 1957 solution video

video by MIT OCW

$$\sqrt{2}/4$$

Convert the integral $$\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }$$ to polar coordinates.

Problem Statement

Convert the integral $$\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }$$ to polar coordinates.

$$\displaystyle{ \int_{\theta=0}^{\pi/4}{ \int_{r=0}^{\tan\theta\sec\theta}{ f~rdr~d\theta } } }$$

Problem Statement

Convert the integral $$\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }$$ to polar coordinates.

Solution

### 1958 solution video

video by MIT OCW

$$\displaystyle{ \int_{\theta=0}^{\pi/4}{ \int_{r=0}^{\tan\theta\sec\theta}{ f~rdr~d\theta } } }$$

Convert $$\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2y-y^2}}{ f~dx~dy } } }$$ to polar coordinates.

Problem Statement

Convert $$\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2y-y^2}}{ f~dx~dy } } }$$ to polar coordinates.

$$\displaystyle{ \int_{\theta=0}^{\pi/2}{ \int_{r=0}^{2\sin\theta}{ f~rdr~d\theta } } }$$

Problem Statement

Convert $$\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2y-y^2}}{ f~dx~dy } } }$$ to polar coordinates.

Solution

### 1959 solution video

video by MIT OCW

$$\displaystyle{ \int_{\theta=0}^{\pi/2}{ \int_{r=0}^{2\sin\theta}{ f~rdr~d\theta } } }$$

$$\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }$$

$$243 \pi/10$$

Problem Statement

$$\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }$$

Solution

### 1960 solution video

video by MIP4U

$$243 \pi/10$$

$$\displaystyle{ \int_{-3}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sin(x^2+y^2)~dy~dx } } }$$

Problem Statement

$$\displaystyle{ \int_{-3}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sin(x^2+y^2)~dy~dx } } }$$

$$(1-\cos 9) \pi/2$$

Problem Statement

$$\displaystyle{ \int_{-3}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sin(x^2+y^2)~dy~dx } } }$$

Solution

### 1961 solution video

video by Krista King Math

$$(1-\cos 9) \pi/2$$

Use a double polar integral to find the volume of the solid enclosed by $$-x^2 - y^2 + z^2 = 1$$ and $$z = 2$$.

Problem Statement

Use a double polar integral to find the volume of the solid enclosed by $$-x^2 - y^2 + z^2 = 1$$ and $$z = 2$$.

$$4 \pi/3$$

Problem Statement

Use a double polar integral to find the volume of the solid enclosed by $$-x^2 - y^2 + z^2 = 1$$ and $$z = 2$$.

Solution

### 1962 solution video

video by Krista King Math

$$4 \pi/3$$

A cylindrical drill with a radius of 5 cm is used to bore a hole through the center of a sphere of radius 7 cm. Find the volume of the ring shaped solid that remains.

Problem Statement

A cylindrical drill with a radius of 5 cm is used to bore a hole through the center of a sphere of radius 7 cm. Find the volume of the ring shaped solid that remains.

$$(4\pi/3)(24)^{3/2}$$ cm3

Problem Statement

A cylindrical drill with a radius of 5 cm is used to bore a hole through the center of a sphere of radius 7 cm. Find the volume of the ring shaped solid that remains.

Solution

### 1967 solution video

video by MIP4U

$$(4\pi/3)(24)^{3/2}$$ cm3

Intermediate Problems

Calculate the area under the plane $$6x + 4y + z = 12$$ above the disk with boundary circle $$x^2 + y^2 = 2y$$.

Problem Statement

Calculate the area under the plane $$6x + 4y + z = 12$$ above the disk with boundary circle $$x^2 + y^2 = 2y$$.

$$8 \pi$$

Problem Statement

Calculate the area under the plane $$6x + 4y + z = 12$$ above the disk with boundary circle $$x^2 + y^2 = 2y$$.

Solution

This problem is solved in three videos. In the first video he sets up the integral. In the second and third videos he evaluates the integral.

### 1955 solution video

video by PatrickJMT

### 1955 solution video

video by PatrickJMT

### 1955 solution video

video by PatrickJMT

$$8 \pi$$

$$\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{x-x^2}}{ xy ~dy ~dx } } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{x-x^2}}{ xy ~dy ~dx } } }$$

$$1/24$$

Problem Statement

$$\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{x-x^2}}{ xy ~dy ~dx } } }$$

Solution

About 2 minutes and 30 seconds into this video he builds what he calls a t-table to try to figure out what boundary of the region looks like. We would probably, instead, try to complete the square to get an equation for the boundary. His way of doing it does not really show that this is a semi-circle of radius $$1/2$$ and center at $$(1/2,0)$$. Here is how we would do it.
We know that $$0\leq y\leq\sqrt{x-x^2}$$. This means that the boundary is $$y=\sqrt{x-x^2}$$ and $$y\geq 0$$. So we complete the square on the boundary equation, as follows.

 $$y = \sqrt{x-x^2}$$ $$y^2 = x-x^2$$ $$0 = x^2-x+y^2$$ $$0 = x^2-x+1/4 -1/4 +y^2$$ $$1/4 = (x-1/2)^2 + y^2$$

In going from line 1 to line 2, we squared both sides of the equation. This is valid for x and y since $$y\geq 0$$.
We would not recommend doing it the way he did in the video, since it is so inexact. However, a graphing calculator or app would help (but don't rely on them too much).

### 1963 solution video

video by MIP4U

$$1/24$$

Find the surface area of the part of the plane $$z = 3 + 2x + 4y$$ that lies inside the cylinder $$x^2 + y^2 = 4$$.

Problem Statement

Find the surface area of the part of the plane $$z = 3 + 2x + 4y$$ that lies inside the cylinder $$x^2 + y^2 = 4$$.

$$4 \pi \sqrt{21}$$

Problem Statement

Find the surface area of the part of the plane $$z = 3 + 2x + 4y$$ that lies inside the cylinder $$x^2 + y^2 = 4$$.

Solution

### 1964 solution video

video by MIP4U

$$4 \pi \sqrt{21}$$

Determine the area of one petal of $$r = 2 \sin(3\theta)$$.

Problem Statement

Determine the area of one petal of $$r = 2 \sin(3\theta)$$.

$$\pi/3$$

Problem Statement

Determine the area of one petal of $$r = 2 \sin(3\theta)$$.

Solution

### 1965 solution video

video by MIP4U

$$\pi/3$$

Find the volume of the solid bounded by the paraboloids $$z = -6 + 3x^2 + 3y^2$$ and $$z = 7 - 2x^2 - 2y^2$$.

Problem Statement

Find the volume of the solid bounded by the paraboloids $$z = -6 + 3x^2 + 3y^2$$ and $$z = 7 - 2x^2 - 2y^2$$.

$$169 \pi/10$$

Problem Statement

Find the volume of the solid bounded by the paraboloids $$z = -6 + 3x^2 + 3y^2$$ and $$z = 7 - 2x^2 - 2y^2$$.

Solution

### 1966 solution video

video by MIP4U

$$169 \pi/10$$