Area Of Circle Double Integral

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Section 4-iv : Double Integrals in Polar Coordinates

To this point we've seen quite a few double integrals. However, in every example we've seen to this point the region \(D\) could be easily described in terms of simple functions in Cartesian coordinates. In this section we want to look at some regions that are much easier to describe in terms of polar coordinates. For instance, we might have a region that is a deejay, ring, or a portion of a disk or ring. In these cases, using Cartesian coordinates could be somewhat cumbersome. For instance, let's suppose nosotros wanted to do the following integral,

\[\iint\limits_{D}{{f\left( {x,y} \correct)\,dA}},\,\,\,\,\,D{\mbox{ is the disk of radius 2}}\]

To this we would have to make up one's mind a set of inequalities for \(x\) and \(y\) that describe this region. These would be,

\[\begin{array}{c} - two \le ten \le 2\\ - \sqrt {4 - {x^2}} \le y \le \sqrt {four - {x^two}} \end{array}\]

With these limits the integral would become,

\[\iint\limits_{D}{{f\left( {x,y} \correct)\,dA}} = \int_{{\, - ii}}^{{\,ii}}{{\int_{{ - \sqrt {4 - {10^2}} }}^{{\,\sqrt {4 - {ten^ii}} }}{{f\left( {ten,y} \right)\,dy}}\,dx}}\]

Due to the limits on the inner integral this is liable to exist an unpleasant integral to compute.

However, a deejay of radius 2 can exist defined in polar coordinates by the post-obit inequalities,

\[\begin{assortment}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\stop{array}\]

These are very simple limits and, in fact, are constant limits of integration which virtually always makes integrals somewhat easier.

So, if nosotros could convert our double integral formula into one involving polar coordinates we would be in pretty skillful shape. The problem is that we tin't just convert the \(dx\) and the \(dy\) into a \(dr\) and a \(d\theta \). In computing double integrals to this bespeak we have been using the fact that \(dA = dx\,dy\) and this really does require Cartesian coordinates to use. One time we've moved into polar coordinates \(dA \ne dr\,d\theta \) so we're going to demand to determine just what \(dA\) is nether polar coordinates.

Then, let'south step dorsum a picayune bit and outset off with a general region in terms of polar coordinates and run into what we can do with that. Here is a sketch of some region using polar coordinates.

So, our general region will exist defined by inequalities,

\[\brainstorm{array}{c}\alpha \le \theta \le \beta \\ {h_1}\left( \theta \correct) \le r \le {h_2}\left( \theta \right)\end{assortment}\]

Now, to find \(dA\) let's redo the figure above as follows,

As shown, we'll intermission upwards the region into a mesh of radial lines and arcs. Now, if we pull one of the pieces of the mesh out equally shown we have something that is almost, but not quite a rectangle. The area of this slice is \(\Delta A\). The two sides of this slice both have length \(\Delta \,r = {r_o} - {r_i}\) where \({r_o}\) is the radius of the outer arc and \({r_i}\) is the radius of the inner arc. Basic geometry and so tells u.s. that the length of the inner edge is \({r_i}\,\Delta \,\theta \) while the length of the out edge is \({r_o}\,\Delta \,\theta \) where \(\Delta \,\theta \) is the angle between the two radial lines that form the sides of this piece.

At present, let'southward assume that we've taken the mesh so modest that nosotros tin assume that \({r_i} \approx {r_o} = r\) and with this supposition we can too assume that our piece is shut enough to a rectangle that nosotros can also then assume that,

\[\Delta A \approx r\,\Delta \,\theta \,\Delta \,r\]

Likewise, if nosotros assume that the mesh is small-scale enough then we can also assume that,

\[dA \approx \Delta A\hspace{0.5in}d\theta \approx \Delta \theta \hspace{0.5in}dr \approx \Delta \,r\]

With these assumptions nosotros then go \(dA \approx r\,dr\,d\theta \).

In order to make it at this we had to make the supposition that the mesh was very small. This is not an unreasonable assumption. Think that the definition of a double integral is in terms of two limits and as limits go to infinity the mesh size of the region will go smaller and smaller. In fact, equally the mesh size gets smaller and smaller the formula above becomes more than and more authentic and and then we tin say that,

\[dA = r\,dr\,d\theta \]

We'll run into another way of deriving this one time nosotros reach the Change of Variables section subsequently in this affiliate. This 2nd mode will non involve any assumptions either and then it maybe a little better way of deriving this.

Earlier moving on it is again of import to annotation that \(dA \ne dr\,d\theta \). The actual formula for \(dA\) has an \(r\) in it. It will exist piece of cake to forget this \(r\) on occasion, but equally y'all'll see without it some integrals volition not be possible to do.

At present, if we're going to be converting an integral in Cartesian coordinates into an integral in polar coordinates we are going to accept to make sure that we've also converted all the \(x\)'s and \(y\)'south into polar coordinates also. To do this nosotros'll need to retrieve the following conversion formulas,

\[x = r\cos \theta \hspace{0.5in}y = r\sin \theta \hspace{0.5in}{r^two} = {10^2} + {y^2}\]

Nosotros are at present ready to write down a formula for the double integral in terms of polar coordinates.

\[\iint\limits_{D}{{f\left( {x,y} \right)\,dA}} = \int_{{\,\alpha }}^{{\,\beta }}{{\int_{{\,{h_{\,1}}\left( \theta \right)}}^{{\,{h_{\,2}}\left( \theta \right)}}{{f\left( {r\cos \theta ,r\sin \theta } \correct)\,r\,dr\,d\theta }}}}\]

It is of import to not forget the added \(r\) and don't forget to convert the Cartesian coordinates in the function over to polar coordinates.

Let'southward await at a couple of examples of these kinds of integrals.

Example ane Evaluate the following integrals by converting them into polar coordinates.

1. \(\displaystyle \iint\limits_{D}{{2x\,y\,dA}}\), \(D\) is the portion of the region between the circles of radius ii and radius 5 centered at the origin that lies in the first quadrant.
2. \(\displaystyle \iint\limits_{D}{{{{\bf{east}}^{{ten^2} + {y^2}}}\,dA}}\), \(D\) is the unit disk centered at the origin.

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a \(\displaystyle \iint\limits_{D}{{2x\,y\,dA}}\), \(D\) is the portion of the region between the circles of radius 2 and radius 5 centered at the origin that lies in the commencement quadrant. Testify Solution

Outset let's go \(D\) in terms of polar coordinates. The circle of radius 2 is given by \(r = 2\) and the circle of radius 5 is given by \(r = 5\). We want the region betwixt the ii circles, so nosotros volition accept the following inequality for \(r\).

\[2 \le r \le five\]

Also, since we simply want the portion that is in the offset quadrant we get the following range of \(\theta \)'s.

\[0 \le \theta \le \frac{\pi }{ii}\]

Now that we've got these we can practise the integral.

\[\iint\limits_{D}{{2x\,y\,dA}} = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\int_{{\,2}}^{{\,five}}{{ii\left( {r\cos \theta } \right)\left( {r\sin \theta } \right)r\,dr}}\,d\theta }}\]

Don't forget to exercise the conversions and to add in the extra \(r\). Now, allow'south simplify and make use of the double angle formula for sine to brand the integral a little easier.

\[\begin{marshal*}\iint\limits_{D}{{2x\,y\,dA}} & = \int_{{\,0}}^{{\,\frac{\pi }{two}}}{{\int_{{\,ii}}^{{\,5}}{{{r^iii}\sin \left( {two\theta } \correct)\,dr}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\left. {\frac{1}{4}{r^4}\sin \left( {ii\theta } \right)} \correct|_2^5\,d\theta }}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{ii}}}{{\frac{{609}}{4}\sin \left( {ii\theta } \right)\,d\theta }}\\ & = \left. { - \frac{{609}}{eight}\cos \left( {2\theta } \correct)} \right|_0^{\frac{\pi }{2}}\\ & = \frac{{609}}{4}\terminate{marshal*}\]

b \(\displaystyle \iint\limits_{D}{{{{\bf{e}}^{{x^2} + {y^two}}}\,dA}}\), \(D\) is the unit of measurement disk centered at the origin. Evidence Solution

In this case we tin't do this integral in terms of Cartesian coordinates. We volition withal be able to practice information technology in polar coordinates. First, the region \(D\) is divers by,

\[\brainstorm{array}{c}0 \le \theta \le ii\pi \\ 0 \le r \le 1\end{assortment}\]

In terms of polar coordinates the integral is then,

\[\iint\limits_{D}{{{{\bf{e}}^{{x^2} + {y^2}}}\,dA}} = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,1}}{{r\,{{\bf{e}}^{{r^2}}}\,dr}}\,d\theta }}\]

Notice that the addition of the \(r\) gives u.s. an integral that nosotros tin now exercise. Here is the work for this integral.

\[\begin{align*}\iint\limits_{D}{{{{\bf{due east}}^{{x^ii} + {y^2}}}\,dA}} & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,one}}{{r\,{{\bf{e}}^{{r^ii}}}\,dr}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\left. {\frac{ane}{two}{{\bf{e}}^{{r^2}}}} \right|_0^1\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{ane}{2}\left( {{\bf{e}} - 1} \right)\,d\theta }}\\ & = \pi \left( {{\bf{e}} - 1} \right)\end{align*}\]

Permit's not forget that we still have the two geometric interpretations for these integrals besides.

Example ii Make up one's mind the expanse of the region that lies inside \(r = three + 2\sin \theta \) and outside \(r = 2\).

Show Solution

Here is a sketch of the region, \(D\), that nosotros want to make up one's mind the area of.

To determine this area we'll need to know that value of \(\theta\) for which the two curves intersect. We can determine these points by setting the 2 equations equal and solving.

\[\begin{marshal*}3 + 2\sin \theta & = two\\ \sin \theta & = - \frac{1}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}\theta = \frac{{seven\pi }}{6},\frac{{eleven\pi }}{vi}\end{align*}\]

Here is a sketch of the figure with these angles added.

Notation as well that we've acknowledged that \( - \frac{\pi }{6}\) is some other representation for the angle \(\frac{{11\pi }}{half-dozen}\). This is important since we need the range of \(\theta\) to actually enclose the regions equally we increase from the lower limit to the upper limit. If we'd chosen to utilize \(\frac{{11\pi }}{6}\) and then as we increase from \(\frac{{vii\pi }}{6}\) to \(\frac{{xi\pi }}{half dozen}\) we would exist tracing out the lower portion of the circle and that is not the region that we are later.

So, hither are the ranges that will define the region.

\[\brainstorm{array}{c}\displaystyle - \frac{\pi }{vi} \le \theta \le \frac{{7\pi }}{vi}\\ 2 \le r \le 3 + 2\sin \theta \end{assortment}\]

To get the ranges for \(r\) the function that is closest to the origin is the lower jump and the role that is farthest from the origin is the upper bound.

The surface area of the region \(D\) is then,

\[\begin{marshal*}A &= \iint\limits_{D}{{dA}}\\ & = \int_{{ - {\pi }/{6}\;}}^{{\,7{\pi }/{half dozen}\;}}{{\int_{2}^{{3 + 2\sin \theta }}{{r\,drd\theta }}}}\\ & = \int_{{ - {\pi }/{6}\;}}^{{\,7{\pi }/{half-dozen}\;}}{{\left. {\frac{1}{two}{r^2}} \right|_2^{3 + 2\sin \theta }\,d\theta }}\\ & = \int_{{ - {\pi }/{half-dozen}\;}}^{{\,vii{\pi }/{6}\;}}{{\frac{5}{2} + six\sin \theta + two{{\sin }^2}\theta \,d\theta }}\\ & = \int_{{ - {\pi }/{half-dozen}\;}}^{{\,seven{\pi }/{6}\;}}{{\frac{7}{ii} + half dozen\sin \theta - \cos \left( {2\theta } \correct)\,d\theta }}\\ & = \left. {\left( {\frac{7}{ii}\theta - 6\cos \theta - \frac{i}{2}\sin \left( {2\theta } \right)} \right)} \correct|_{ - \frac{\pi }{half-dozen}}^{\frac{{7\pi }}{6}}\\ & = \frac{{11\sqrt 3 }}{ii} + \frac{{fourteen\pi }}{3} = 24.187\stop{align*}\]

Example three Determine the volume of the region that lies under the sphere \({ten^2} + {y^2} + {z^2} = 9\), in a higher place the airplane \(z = 0\) and inside the cylinder \({ten^2} + {y^2} = 5\).

Testify Solution

Nosotros know that the formula for finding the book of a region is,

\[V = \iint\limits_{D}{{f\left( {x,y} \right)\,dA}}\]

In order to brand use of this formula we're going to need to determine the function that we should be integrating and the region \(D\) that nosotros're going to exist integrating over.

The function isn't as well bad. It's just the sphere, nonetheless, we exercise demand information technology to be in the form \(z = f\left( {x,y} \right)\). We are looking at the region that lies under the sphere and in a higher place the plane \(z = 0\) (simply the \(xy\)-plane right?) and so all nosotros need to do is solve the equation for \(z\) and when taking the foursquare root nosotros'll accept the positive i since we are wanting the region above the \(xy\)-aeroplane. Here is the function.

\[z = \sqrt {9 - {x^two} - {y^2}} \]

The region \(D\) isn't too bad in this example either. Equally nosotros take points, \(\left( {x,y} \right)\), from the region we need to completely graph the portion of the sphere that nosotros are working with. Since nosotros only want the portion of the sphere that really lies inside the cylinder given by \({x^2} + {y^2} = 5\) this is also the region \(D\). The region \(D\) is the disk \({x^ii} + {y^2} \le 5\) in the \(xy\)-airplane.

For reference purposes here is a sketch of the region that we are trying to find the volume of.

So, the region that nosotros want the volume for is really a cylinder with a cap that comes from the sphere.

We are definitely going to want to practise this integral in terms of polar coordinates so here are the limits (in polar coordinates) for the region,

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le \sqrt 5 \cease{array}\]

and we'll need to convert the function to polar coordinates besides.

\[z = \sqrt {9 - \left( {{x^2} + {y^2}} \right)} = \sqrt {9 - {r^ii}} \]

The volume is then,

\[\begin{align*}V & = \iint\limits_{D}{{\sqrt {nine - {x^2} - {y^2}} \,dA}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,\sqrt five }}{{r\sqrt {9 - {r^ii}} \,dr}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{ - \frac{1}{3}\left. {{{\left( {ix - {r^2}} \right)}^{\frac{3}{2}}}} \right|_0^{\sqrt 5 }\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{{19}}{3}\,d\theta }}\\ & = \frac{{38\pi }}{three}\end{align*}\]

Case four Observe the volume of the region that lies inside \(z = {x^ii} + {y^two}\) and below the aeroplane \(z = xvi\).

Show Solution

Permit's start this example off with a quick sketch of the region.

Now, in this example the standard formula is not going to work. The formula

\[Five = \iint\limits_{D}{{f\left( {x,y} \right)\,dA}}\]

finds the volume under the function \(f\left( {x,y} \right)\) and nosotros're actually after the volume that is higher up a office. This isn't the problem that it might appear to be however. First, discover that

\[Five = \iint\limits_{D}{{16\,dA}}\]

will be the volume under \(z = 16\) (of course we'll need to determine \(D\) somewhen) while

\[5 = \iint\limits_{D}{{{x^ii} + {y^two}\,dA}}\]

is the volume under \(z = {x^2} + {y^2}\), using the aforementioned \(D\).

The volume that we're after is really the divergence betwixt these 2 or,

\[Five = \iint\limits_{D}{{16\,dA}} - \iint\limits_{D}{{{ten^two} + {y^2}\,dA}} = \iint\limits_{D}{{16 - \left( {{ten^2} + {y^2}} \right)\,dA}}\]

Now all that we demand to practice is to determine the region \(D\) then catechumen everything over to polar coordinates.

Determining the region \(D\) in this case is non too bad. If we were to look direct down the \(z\)-axis onto the region we would come across a circle of radius iv centered at the origin. This is because the meridian of the region, where the elliptic paraboloid intersects the airplane, is the widest function of the region. We know the \(z\) coordinate at the intersection then, setting \(z = sixteen\) in the equation of the paraboloid gives,

\[16 = {x^2} + {y^two}\]

which is the equation of a circumvolve of radius iv centered at the origin.

Here are the inequalities for the region and the role nosotros'll be integrating in terms of polar coordinates.

\[0 \le \theta \le 2\pi \hspace{0.5in}0 \le r \le 4\hspace{0.5in}z = 16 - {r^2}\]

The book is and so,

\[\begin{marshal*}V & = \iint\limits_{D}{{16 - \left( {{ten^2} + {y^2}} \correct)\,dA}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,4}}{{r\left( {16 - {r^2}} \right)\,dr}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,two\pi }}{{\left. {\left( {8{r^2} - \frac{1}{4}{r^four}} \right)} \correct|}}_0^4\,d\theta \\ & = \int_{{\,0}}^{{\,2\pi }}{{64\,d\theta }}\,\\ & = 128\pi \end{align*}\]

In both of the previous volume problems we would have not been able to easily compute the volume without first converting to polar coordinates then, as these examples show, it is a good idea to always recall polar coordinates.

There is i more than blazon of instance that nosotros need to look at earlier moving on to the next section. Sometimes we are given an iterated integral that is already in terms of \(x\) and \(y\) and we need to catechumen this over to polar then that we tin can actually exercise the integral. We need to see an example of how to practice this kind of conversion.

Example 5 Evaluate the following integral by first converting to polar coordinates.

\[\int_{{ - 1}}^{1}{{\int_{{ - \sqrt {one - {x^2}} }}^{0}{{\cos \left( {{ten^ii} + {y^2}} \right)\,dy}}\,dx}}\]

Bear witness Solution

Showtime, detect that we cannot do this integral in Cartesian coordinates then converting to polar coordinates may exist the only option we accept for actually doing the integral. Notice that the part will catechumen to polar coordinates nicely then shouldn't be a problem.

Permit's first decide the region that nosotros're integrating over and see if information technology'south a region that can be easily converted into polar coordinates. Here are the inequalities that define the region in terms of Cartesian coordinates.

\[\begin{array}{c} - 1 \le x \le one\\ - \sqrt {1 - {x^ii}} \le y \le 0\end{assortment}\]

Now, the lower limit for the \(y\)'s is,

\[y = - \sqrt {1 - {x^ii}} \]

and this looks like the bottom of the circle of radius 1 centered at the origin. Since the upper limit for the \(y\)'s is \(y = 0\) we won't have whatever portion of the meridian half of the disk and and then information technology looks like we are going to have a portion (or all) of the lesser of the disk of radius 1 centered at the origin.

The range for the \(10\)'s in plow, tells u.s. that nosotros are volition in fact have the complete bottom part of the disk.

So, nosotros know that the inequalities that will define this region in terms of polar coordinates are then,

\[\begin{array}{c}\pi \le \theta \le 2\pi \\ 0 \le r \le 1\end{array}\]

Finally, we simply need to remember that,

\[dx\,dy = dA = r\,dr\,d\theta \]

and and so the integral becomes,

\[\int_{{ - 1}}^{1}{{\int_{{ - \sqrt {1 - {x^two}} }}^{0}{{\cos \left( {{x^2} + {y^two}} \right)\,dy}}\,dx}} = \int_{\pi }^{{2\pi }}{{\int_{0}^{1}{{r\cos \left( {{r^ii}} \correct)\,dr}}\,d\theta }}\]

Annotation that this is an integral that we tin do. And then, here is the residuum of the work for this integral.

\[\begin{align*}\int_{{ - 1}}^{ane}{{\int_{{ - \sqrt {1 - {x^2}} }}^{0}{{\cos \left( {{ten^two} + {y^2}} \correct)\,dy}}\,dx}} & = \int_{\pi }^{{2\pi }}{{\left. {\frac{1}{2}\sin \left( {{r^2}} \right)} \right|_0^one\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\frac{1}{2}\sin \left( i \right)\,d\theta }}\\ & = \frac{\pi }{2}\sin \left( 1 \right)\stop{align*}\]

Area Of Circle Double Integral,

Source: https://tutorial.math.lamar.edu/classes/calciii/dipolarcoords.aspx

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