Wize University Calculus 3 Textbook > Multiple Integrals

The polar coordinate system is a substitute to the Cartesian xy-coordinate system. In this system, rather than specifying the coordinate of a point with x and y coordinates in 2D space, we can use the radius and the angle to define the coordinate of a point uniquely. The radius and angle with respect to the positive x-axis are shown below:
The following equations relates the polar coordinates system with Cartesian coordinates system:
x=rcos⁡θy=rsin⁡θx2+y2=(rcos⁡θ)2+(rsin⁡θ)2=r2\begin{array}{c}
x=r\cos\theta\\[10pt]
y=r\sin\theta\\[10pt]
x^2+y^2=(r\cos\theta)^2+(r\sin\theta)^2= r^2
\end{array}x=rcosθy=rsinθx2+y2=(rcosθ)2+(rsinθ)2=r2​

Wize Concept
When we convert the area to polar coordinates, we will have dA=dr⋅rdθdA=dr\cdot rd\thetadA=dr⋅rdθ

The general form of a double integral from rectangular coordinates to polar coordinates is in the form of:
∬Df(x,y)dA=∫αβ∫abf(rcos⁡θ,rsin⁡θ)rdrdθ\iint_Df(x,y)dA=\int_\alpha^\beta\int_a^bf(r\cos\theta,r\sin\theta)rdrd\theta∬D​f(x,y)dA=∫αβ​∫ab​f(rcosθ,rsinθ)rdrdθ
where r changes from a to b and θ\thetaθ changes from α\alphaα to β\betaβ.

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Evaluate the double integral
I=∫∫Dsin⁡(x2+y2)dAI=\int_{ }^{ }\int_D^{ }\sin(x^2+y^2)dAI=∫​∫D​sin(x2+y2)dA
where DDD is a region inside a circle with a radius of 3 in the first quadrant.

I=∬Dsin⁡(x2+y2)dA                                                =∫0π2∫03sin⁡(r2)rdrdθ       Let u=r2. Then, du2=rdr         =12∫0π2∫03(sin⁡u du)dθ        =12∫0π2dθ ⋅ (−cos⁡u)∣03        =12∫0π2dθ ⋅ (−cos⁡r2)∣03             =12∫0π2dθ ⋅ (−cos⁡9+1)  =12(1−cos⁡9)∫0π2dθ=12(1−cos⁡9)∣0π2     =π4(1−cos⁡9)       \displaystyle I=\iint_{D}\sin(x^2+y^2)dA\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\int_0^{\frac{\pi}{2}}\int_0^3\sin(r^2)rdrd\theta~~~~{\scriptsize\color{magenta}~~~Let~u=r^2.~Then,~\frac{du}{2}=rdr}\newline{}\newline{}
~~~~~~~~~
=\frac{1}{2}\int_0^{\frac{\pi}{2}}\int_0^3\Big(\sin{u}~du\Big)d\theta\newline{}\newline{}
~~~~~~~~
=\frac{1}{2}\int_0^{\frac{\pi}{2}}d\theta~\cdot~(-\cos{u})\Big|_0^3\newline{}\newline{}
~~~~~~~~
=\frac{1}{2}\int_0^{\frac{\pi}{2}}d\theta~\cdot~(-\cos{r^2})\Big|_0^3\newline{}\newline{}
~~~~~~~~~~~~~
=\frac{1}{2}\int_0^{\frac{\pi}{2}}d\theta~\cdot~\Big(-\cos{9}+1\Big)\newline{}\newline{}
~~
=\frac{1}{2}(1-\cos9)\int_0^{\frac{\pi}{2}}d\theta\newline{}\newline{}

=\frac{1}{2}(1-\cos9)\Big|_0^{\frac{\pi}{2}}~~~~~\newline{}\newline{}

=\frac{\pi}{4}(1-\cos9)~~~~~~~I=∬D​sin(x2+y2)dA                                                =∫02π​​∫03​sin(r2)rdrdθ       Let u=r2. Then, 2du​=rdr         =21​∫02π​​∫03​(sinu du)dθ        =21​∫02π​​dθ ⋅ (−cosu)​03​        =21​∫02π​​dθ ⋅ (−cosr2)​03​             =21​∫02π​​dθ ⋅ (−cos9+1)  =21​(1−cos9)∫02π​​dθ=21​(1−cos9)​02π​​     =4π​(1−cos9)       

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Evaluate ∫D∫2xy dA, \displaystyle\int_D\int{}2xy~dA,~∫D​∫2xy dA, where D is the portion of the region between the circles of radius 2 and radius 5 centered at the origin that lies in the first quadrant. 

We can determine the domain, D, in terms of polar coordinates. So,
2≤r≤5,  0≤θ≤π22\leq{r}\leq{5},~~0\leq{\theta}\leq{\frac{\pi}{2}}2≤r≤5,  0≤θ≤2π​
Then, 
∫D∫2xy dA=∫0π2∫252(rcosθ)(rsinθ)rdrdθ         =∫0π2∫25r3sin⁡2θ drdθ              =∫0π2sin⁡2θ dθ ⋅ (r44)∣25                      =∫0π2sin⁡2θ dθ ⋅ (6254 − 4)  =−6098 ⋅ (cos2θ)∣0π2                       =−6098[cos⁡(2π2) − cos⁡(0)]=6094                   \displaystyle\int_D\int2xy~dA=\int_{0}^{\frac{\pi}{2}}\int_{2}^{5}2(rcos\theta)(rsin\theta)rdrd\theta\newline{}\newline{}
~~~~~~~~~
=\int_{0}^{\frac{\pi}{2}}\int_{2}^{5}r^3\sin2\theta~drd\theta\newline{}\newline{}
~~~~~~~~~~~~~~
=\int_{0}^{\frac{\pi}{2}}\sin2\theta~d\theta~\cdot~\Bigg(\frac{r^4}{4}\Bigg)\Bigg|_2^5\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~
=\int_{0}^{\frac{\pi}{2}}\sin2\theta~d\theta~\cdot~\Bigg(\frac{625}{4}~-~4\Bigg)\newline{}\newline{}
~~
=\frac{-609}{8}~\cdot~\big(cos2\theta\big)\bigg|_0^{\frac{\pi}{2}}\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~
=\frac{-609}{8}\Bigg[\cos\bigg({2\frac{\pi}{2}}\bigg)~-~\cos(0)
\Bigg]\newline{}\newline{}

=\frac{609}{4}~~~~~~~~~~~~~~~~~~~∫D​∫2xy dA=∫02π​​∫25​2(rcosθ)(rsinθ)rdrdθ         =∫02π​​∫25​r3sin2θ drdθ              =∫02π​​sin2θ dθ ⋅ (4r4​)​25​                      =∫02π​​sin2θ dθ ⋅ (4625​ − 4)  =8−609​ ⋅ (cos2θ)​02π​​                       =8−609​[cos(22π​) − cos(0)]=4609​                   

What is the volume of the region under the sphere x2+y2+z2=9, x^2+y^2+z^2=9,~x2+y2+z2=9, above the plane z=0, z=0,~z=0, and inside  x2+y2=5?~x^2+y^2=5? x2+y2=5?

Answer

I don't know

Find ∫D∫1+4x2+4y2dA \displaystyle\int_{D}^{}\int_{}^{}\sqrt{1+4x^2+4y^2}dA~∫D​∫​1+4x2+4y2​dA where D is the lower half of x2+y2=16x^2+y^2=16x2+y2=16

Answer

I don't know