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Double integrals are sometimes much easier to advice if we readjust rectangular collaborates to polar coordinates. However, before we describe how to do this change, we need to create the concept of a double integral in a polar rectangle-shaped region.
Polar rectangular Regions that Integration
When we defined the twin integral for a consistent function in rectangle-shaped coordinates—say, \(g\) end a region \(R\) in the \(xy\)-plane—we separated \(R\) into subrectangles with sides parallel come the name: coordinates axes. This sides have actually either continuous \(x\)-values and/or consistent \(y\)-values. In polar coordinates, the form we job-related with is a polar rectangle, whose sides have constant \(r\)-values and/or consistent \(\theta\)-values. This method we can define a polar rectangle as in figure \(\PageIndex1a\), through \(R = \(r,\theta)\,\).
In this section, we are looking to integrate over polar rectangles. Think about a role \(f(r,\theta)\) over a polar rectangle \(R\). We division the term \(\) into \(m\) subintervals \(
As before, we need to find the area \(\Delta A\) of the polar subrectangle \(R_ij\) and the “polar” volume that the slim box above \(R_ij\). Recall that, in a one of radius \(r\) the size \(s\) of an arc subtended by a main angle the \(\theta\) radians is \(s = r\theta\). Notice that the polar rectangle \(R_ij\) looks a lot choose a trapezoid through parallel sides \(r_i-1\Delta \theta\) and \(r_i\Delta \theta\) and with a width \(\Delta r\). Thus the area that the polar subrectangle \(R_ij\) is
\<\Delta A = \frac12 \Delta r (r_i-1 \Delta \theta + r_i \Delta \theta ). \nonumber\>
Simplifying and also letting
we have \(\Delta A = r_ij^* \Delta r \Delta \theta\).
Therefore, the polar volume that the slim box over \(R_ij\) (Figure \(\PageIndex2\)) is
Using the very same idea for all the subrectangles and summing the volumes of the rectangle-shaped boxes, we obtain a dual Riemann sum as
\<\sum_i=1^m \sum_j=1^n f(r_ij^*, \theta_ij^*) r_ij^* \Delta r \Delta \theta. \nonumber\>
As we have actually seen before, we achieve a much better approximation to the polar volume that the solid above the an ar \(R\) as soon as we let \(m\) and also \(n\) become larger. Hence, we specify the polar volume together the limit of the twin Riemann sum,
This becomes the expression because that the dual integral.
Again, simply as in ar on double Integrals over rectangle-shaped Regions, the double integral over a polar rectangular an ar can it is in expressed together an iterated integral in polar coordinates. Hence,
\<\iint_R f(r, \theta)\,dA = \iint_R f(r, \theta) \,r \, dr \, d\theta = \int_\theta=\alpha^\theta=\beta \int_r=a^r=b f(r,\theta) \,r \, dr \, d\theta.\>
Notice the the expression because that \(dA\) is replaced by \(r \, dr \, d\theta\) as soon as working in polar coordinates. Another method to look at the polar double integral is to adjust the dual integral in rectangular collaborates by substitution. As soon as the duty \(f\) is provided in terms of \(x\) and \(y\) utilizing \(x = r \, \cos \, \theta, \, y = r \, \sin \, \theta\), and \(dA = r \, dr \, d\theta\) transforms it to
\<\iint_R f(x,y) \,dA = \iint_R f(r \, \cos \, \theta, \, r \, \sin \, \theta ) \,r \, dr \, d\theta.\>
Note that all the properties listed in section on double Integrals over rectangle-shaped Regions because that the twin integral in rectangular collaborates hold true for the double integral in polar works with as well, so we deserve to use them without hesitation.
Example \(\PageIndex1A\): Sketching a Polar rectangle-shaped Region
Sketch the polar rectangular region
As we can see from number \(\PageIndex3\), \(r = 1\) and \(r = 3\) room circles that radius 1 and 3 and \(0 \leq \theta \leq \pi\) covers the entire top half of the plane. For this reason the region \(R\) looks favor a semicircular band.
Now that we have actually sketched a polar rectangle-shaped region, allow us show how to advice a twin integral end this an ar by using polar coordinates.
Example \(\PageIndex1B\): examining a dual Integral end a Polar rectangular Region
Evaluate the integral \(\displaystyle \iint_R 3x \, dA\) over the region \(R = \(r, \theta)\,.\)
First we sketch a figure comparable to number \(\PageIndex3\), yet with external radius \(r=2\). From the number we can see that we have
\<\beginalign* \iint_R 3x \, dA &= \int_\theta=0^\theta=\pi \int_r=1^r=2 3r \, \cos \, \theta \,r \, dr \, d\theta \quad\textUse an integral v correct limits of integration. \\ &= \int_\theta=0^\theta=\pi \cos \, \theta \left<\left. R^3\right|_r=1^r=2\right> d\theta \quad\textIntegrate an initial with respect come $r$. \\ &=\int_\theta=0^\theta=\pi 7 \, \cos \, \theta \, d\theta \\ &= 7 \, \sin \, \theta \bigg|_\theta=0^\theta=\pi = 0. \endalign*\>
Sketch the an ar \(D = \ (r,\theta) \vert 1\leq r \leq 2, \, -\frac\pi2 \leq \theta \leq \frac\pi2 \\), and evaluate \(\displaystyle \iint_R x \, dA\).Hint
Follow the measures in instance \(\PageIndex1A\).Answer
Example \(\PageIndex2A\): analyzing a double Integral by convert from rectangle-shaped Coordinates
Evaluate the integral
\<\iint_R (1 - x^2 - y^2) \,dA \nonumber\>
where \(R\) is the unit one on the \(xy\)-plane.
The an ar \(R\) is a unit circle, so we can describe it together \(R = \\,0 \leq r \leq 1, \, 0 \leq \theta \leq 2\pi \\).
Using the conversion \(x = r \, \cos \, \theta, \, y = r \, \sin \, \theta\), and \(dA = r \, dr \, d\theta\), us have
\<\beginalign* \iint_R (1 - x^2 - y^2) \,dA &= \int_0^2\pi \int_0^1 (1 - r^2) \,r \, dr \, d\theta \\<4pt> &= \int_0^2\pi \int_0^1 (r - r^3) \,dr \, d\theta \\ &= \int_0^2\pi \left<\fracr^22 - \fracr^44\right>_0^1 \,d\theta \\&= \int_0^2\pi \frac14\,d\theta = \frac\pi2. \endalign*\>
Example \(\PageIndex2B\): evaluating a twin Integral by convert from rectangle-shaped Coordinates
Evaluate the integral \<\displaystyle \iint_R (x + y) \,dA \nonumber\> where \(R = \big\(x,y)\,.\)
We have the right to see the \(R\) is one annular an ar that deserve to be converted to polar coordinates and described as \(R = \left\\,1 \leq r \leq 2, \, \frac\pi2 \leq \theta \leq \frac3\pi2 \right\\) (see the following graph).
Hence, using the switch \(x = r \, \cos \, \theta, \, y = r \, \sin \, \theta\), and also \(dA = r \, dr \, d\theta\), we have
\<\beginalign* \iint_R (x + y)\,dA &= \int_\theta=\pi/2^\theta=3\pi/2 \int_r=1^r=2 (r \, \cos \, \theta + r \, \sin \, \theta) r \, dr \, d\theta \\ &= \left(\int_r=1^r=2 r^2 \, dr\right)\left(\int_\pi/2^3\pi/2 (\cos \, \theta + \sin \, \theta)\,d\theta\right) \\ &= \left. \left<\fracr^33\right>_1^2 <\sin \, \theta - \cos \, \theta> \right|_\pi/2^3\pi/2 \\ &= - \frac143. \endalign*\>
General Polar areas of Integration
To advice the twin integral of a consistent function through iterated integrals over basic polar regions, we take into consideration two varieties of regions, analogous to form I and form II as discussed for rectangular works with in section on dual Integrals over basic Regions. It is an ext common to compose polar equations together \(r = f(\theta)\) than \(\theta = f(r)\), so we explain a general polar region as \(R = \(r, \theta)\,\) (Figure \(\PageIndex5\)).
Hence, we have
\<\beginalign* \iint_D r^2 \sin \, \theta \, r \, dr \, d\theta &= \int_\theta=0^\theta=\pi \int_r=0^r=1+\cos \theta (r^2 \sin \, \theta) \,r \, dr \, d\theta \\ &= \frac14\left.\int_\theta=0^\theta=\pi
Evaluate the integral
\<\iint_D r^2 \sin^2 2\theta \,r \, dr \, d\theta \nonumber\>
where \(D = \left\ (r,\theta)\,\).Hint
Graph the an ar and monitor the steps in the ahead example.Answer
Polar Areas and also Volumes
As in rectangle-shaped coordinates, if a solid \(S\) is bounded by the surface \(z = f(r, \theta)\), as well as by the surface \(r = a, \, r = b, \, \theta = \alpha\), and \(\theta = \beta\), we can discover the volume \(V\) of \(S\) by twin integration, as
If the base of the solid can be defined as \(D = \(r, \theta)\), then the twin integral for the volume becomes
We show this idea through some examples.
Example \(\PageIndex4A\): recognize a Volume using a dual Integral
Find the volume the the solid that lies under the paraboloid \(z = 1 - x^2 - y^2\) and over the unit one on the \(xy\)-plane (Figure \(\PageIndex7\)).
By the technique of dual integration, we can see the the volume is the iterated integral of the form
\<\displaystyle \iint_R (1 - x^2 - y^2)\,dA \nonumber\>
where \(R = \big\\,0 \leq r \leq 1, \, 0 \leq \theta \leq 2\pi\big\\).
This integration was shown before in example \(\PageIndex2A\), for this reason the volume is \(\frac\pi2\) cubic units.
Example \(\PageIndex4B\): finding a Volume Using dual Integration
Find the volume the the solid that lies under the paraboloid \(z = 4 - x^2 - y^2\) and over the disk \((x - 1)^2 + y^2 = 1\) on the \(xy\)-plane. View the paraboloid in number \(\PageIndex8\) intersecting the cylinder \((x - 1)^2 + y^2 = 1\) above the \(xy\)-plane.
First readjust the decaying \((x - 1)^2 + y^2 = 1\) to polar coordinates. Expanding the square term, we have actually \(x^2 - 2x + 1 + y^2 = 1\). Then simplify to gain \(x^2 + y^2 = 2x\), i beg your pardon in polar collaborates becomes \(r^2 = 2r \, \cos \, \theta\) and then either \(r = 0\) or \(r = 2 \, \cos \, \theta\). Similarly, the equation the the paraboloid changes to \(z = 4 - r^2\). Because of this we can define the disk \((x - 1)^2 + y^2 = 1\) on the \(xy\) -plane together the region
Hence the volume the the hard bounded above by the paraboloid \(z = 4 - x^2 - y^2\) and below by \(r = 2 \, \cos \theta\) is
\<\beginalign* V &= \iint_D f(r, \theta) \,r \, dr \, d\theta \\&= \int_\theta=0^\theta=\pi \int_r=0^r=2 \, \cos \, \theta (4 - r^2) \,r \, dr \, d\theta\\ &= \int_\theta=0^\theta=\pi\left.\left<4\fracr^22 - \fracr^44\right|_0^2 \, \cos \, \theta\right>d\theta \\ &= \int_0^\pi <8 \, \cos^2\theta - 4 \, \cos^4\theta>\,d\theta \\&= \left<\frac52\theta + \frac52 \sin \, \theta \, \cos \, \theta - \sin \, \theta \cos^3\theta \right>_0^\pi = \frac52\pi\; \textunits^3. \endalign*\>
Example \(\PageIndex5A\): recognize a Volume utilizing a twin Integral
Find the volume that the region that lies under the paraboloid \(z = x^2 + y^2\) and over the triangle fastened by the present \(y = x, \, x = 0\), and also \(x + y = 2\) in the \(xy\)-plane.
First study the an ar over i m sorry we require to collection up the twin integral and the accompanying paraboloid.
The region \(D\) is \(\\,0 \leq x \leq 1, \, x \leq y \leq 2 - x\\). Convert the currently \(y = x, \, x = 0\), and \(x + y = 2\) in the \(xy\)-plane to attributes of \(r\) and also \(\theta\) we have actually \(\theta = \pi/4, \, \theta = \pi/2\), and \(r = 2 / (\cos \, \theta + \sin \, \theta)\), respectively. Graphing the an ar on the \(xy\)- plane, we watch that that looks favor \(D = \(r, \theta)\,\).
Now converting the equation that the surface provides \(z = x^2 + y^2 = r^2\). Therefore, the volume of the heavy is given by the dual integral
\<\beginalign* V &= \iint_D f(r, \theta)\,r \, dr \, d\theta \\&= \int_\theta=\pi/4^\theta=\pi/2 \int_r=0^r=2/ (\cos \, \theta + \sin \, \theta) r^2 r \, dr d\theta \\ &= \int_\pi/4^\pi/2\left<\fracr^44\right>_0^2/(\cos \, \theta + \sin \, \theta) d\theta \\ &=\frac14\int_\pi/4^\pi/2 \left(\frac2\cos \, \theta + \sin \, \theta\right)^4 d\theta \\ &= \frac164 \int_\pi/4^\pi/2 \left(\frac1\cos \, \theta + \sin \, \theta \right)^4 d\theta \\&= 4\int_\pi/4^\pi/2 \left(\frac1\cos \, \theta + \sin \, \theta\right)^4 d\theta. \endalign*\>
As you can see, this integral is very complicated. So, we can instead evaluate this twin integral in rectangular collaborates as
\<\beginalign* V &= \int_0^1 \int_x^2-x (x^2 + y^2) \,dy \, dx \\&= \int_0^1 \left.\left
Example \(\PageIndex5B\): detect a Volume making use of a twin Integral
Use polar collaborates to uncover the volume within the cone \(z = 2 - \sqrtx^2 + y^2\) and above the \(xy\)-plane.
The an ar \(D\) for the integration is the basic of the cone, which appears to it is in a circle on the \(xy\)-plane (Figure \(\PageIndex10\)).
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Using symmetry, we can see the we need to discover the area the one petal and also then main point it by 8. Notice that the worths of \(\theta\) for which the graph passes with the beginning are the zeros the the function \(\cos \, 4\theta\), and also these space odd multiples of \(\pi/8\). Thus, one of the petals corresponds to the values of \(\theta\) in the term \(<-\pi/8, \pi/8>\). Therefore, the area bounded through the curve \(r = \cos \, 4\theta\) is
\<\beginalign* A &= 8 \int_\theta=-\pi/8^\theta=\pi/8 \int_r=0^r=\cos \, 4\theta 1\,r \, dr \, d\theta \\ &= 8 \int_\theta=-\pi/8^\theta=\pi/8\left.\left<\frac12r^2\right|_0^\cos \, 4\theta\right> d\theta \\ &= 8 \int_-\pi/8^\pi/8 \frac12 \cos^24\theta \, d\theta \\&= 8\left. \left<\frac14 \theta + \frac116 \sin \, 4\theta \, \cos \, 4\theta \right|_-\pi/8^\pi/8\right> \\&= 8 \left<\frac\pi16\right> = \frac\pi2\; \textunits^2. \endalign*\>
api/deki/files/10973/imageedit_25_4423207893.png?revision=1" />Figure \(\PageIndex12\): finding the area attached by both a circle and a cardioid.
We deserve to from see the the contrary of the graph the we need to discover the clues of intersection. Setup the two equations same to each various other gives
\<3 \, \cos \, \theta = 1 + \cos \, \theta. \nonumber\>
One that the points of intersection is \(\theta = \pi/3\). The area over the polar axis consists of 2 parts, with one part defined through the cardioid native \(\theta = 0\) come \(\theta = \pi/3\) and the other component defined through the circle from \(\theta = \pi/3\) come \(\theta = \pi/2\). By symmetry, the total area is twice the area above the polar axis. Thus, we have