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Area and Arc Length in Polar Coordinates
Apply the formula for area of a region in polar coordinates.
Areas of Regions Bounded by Polar Curves
We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.
Consider a curve defined by the function \(r=f(\theta ),\) where \(\alpha \le \theta \le \beta .\) Our first step is to partition the interval \([\alpha ,\beta ]\) into n equal-width subintervals. The width of each subinterval is given by the formula \(\Delta \theta =(\beta -\alpha )\text{/}n,\) and the ith partition point \({\theta }_{i}\) is given by the formula \({\theta }_{i}=\alpha +i\Delta \theta .\) Each partition point \(\theta ={\theta }_{i}\) defines a line with slope \(\text{tan}{\theta }_{i}\) passing through the pole as shown in the following graph.
The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.
Recall that the area of a circle is \(A=\pi {r}^{2}.\) When measuring angles in radians, 360 degrees is equal to \(2\pi\) radians. Therefore a fraction of a circle can be measured by the central angle \(\theta .\) The fraction of the circle is given by \(\frac{\theta }{2\pi },\) so the area of the sector is this fraction multiplied by the total area:
\[A=(\frac{\theta }{2\pi })\ \pi {r}^{2}=\frac{1}{2}\theta {r}^{2}.\]Since the radius of a typical sector in is given by \({r}_{i}=f({\theta }_{i}),\) the area of the ith sector is given by
\[{A}_{i}=\frac{1}{2}(\Delta \theta ){(f({\theta }_{i}))}^{2}.\]Therefore a Riemann sum that approximates the area is given by
\[{A}_{n}=\sum _{i=1}^{n}{A}_{i}\approx \sum _{i=1}^{n}\frac{1}{2}(\Delta \theta ){(f({\theta }_{i}))}^{2}.\]We take the limit as \(n\to \infty\) to get the exact area:
\[A=\underset{n\to \infty }{\text{lim}}{A}_{n}=\frac{1}{2}{\int }_{\alpha }^{\beta }{(f(\theta ))}^{2}d\theta .\]\[6\ \text{sin}\ \theta =0.\]Condensed: the full section is in OpenStax Calculus Volume 3.
Arc Length in Polar Curves
Here we derive a formula for the arc length of a curve defined in polar coordinates.
In rectangular coordinates, the arc length of a parameterized curve \((x(t),y(t))\) for \(a\le t\le b\) is given by
\[L={\int }_{a}^{b}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt.\]In polar coordinates we define the curve by the equation \(r=f(\theta ),\) where \(\alpha \le \theta \le \beta .\) In order to adapt the arc length formula for a polar curve, we use the equations
\[x=r\ \text{cos}\ \theta =f(\theta )\ \text{cos}\ \theta \ \text{and}\ y=r\ \text{sin}\ \theta =f(\theta )\ \text{sin}\ \theta ,\]and we replace the parameter t by \(\theta .\) Then
\[\begin{array}{l}\frac{dx}{d\theta }={f}^{'}(\theta )\ \text{cos}\ \theta -f(\theta )\ \text{sin}\ \theta \\ \frac{dy}{d\theta }={f}^{'}(\theta )\ \text{sin}\ \theta +f(\theta )\ \text{cos}\ \theta .\end{array}\]We replace \(dt\) by \(d\theta ,\) and the lower and upper limits of integration are \(\alpha\) and \(\beta ,\) respectively. Then the arc length formula becomes
\[\begin{array}{ll}L & ={\int }_{a}^{b}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt \\ & ={\int }_{\alpha }^{\beta }\sqrt{{(\frac{dx}{d\theta })}^{2}+{(\frac{dy}{d\theta })}^{2}}d\theta \\ & ={\int }_{\alpha }^{\beta }\sqrt{{({f}^{'}(\theta )\ \text{cos}\ \theta -f(\theta )\ \text{sin}\ \theta )}^{2}+{({f}^{'}(\theta )\ \text{sin}\ \theta +f(\theta )\ \text{cos}\ \theta )}^{2}}d\theta \\ & ={\int }_{\alpha }^{\beta }\sqrt{{({f}^{'}(\theta ))}^{2}({\text{cos}}^{2}\theta +{\text{sin}}^{2}\theta )+{(f(\theta ))}^{2}({\text{cos}}^{2}\theta +{\text{sin}}^{2}\theta )}d\theta \\ & ={\int }_{\alpha }^{\beta }\sqrt{{({f}^{'}(\theta ))}^{2}+{(f(\theta ))}^{2}}d\theta \\ & ={\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{(\frac{dr}{d\theta })}^{2}}d\theta .\end{array}\]This gives us the following theorem.
Condensed: the full section is in OpenStax Calculus Volume 3.
Key Concepts
- The area of a region in polar coordinates defined by the equation \(r=f(\theta )\) with \(\alpha \le \theta \le \beta\) is given by the integral \(A=\frac{1}{2}{{\int }_{\alpha }^{\beta }[f(\theta )]}^{2}d\theta .\)
- To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
- The arc length of a polar curve defined by the equation \(r=f(\theta )\) with \(\alpha \le \theta \le \beta\) is given by the integral \(L={\int }_{\alpha }^{\beta }\sqrt{{[f(\theta )]}^{2}+{[{f}^{'}(\theta )]}^{2}}d\theta ={\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{(\frac{dr}{d\theta })}^{2}}d\theta .\)
Key Equations
| Area of a region bounded by a polar curve | \(A=\frac{1}{2}{\int }_{\alpha }^{\beta }{[f(\theta )]}^{2}d\theta =\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^{2}d\theta\) |
| Arc length of a polar curve | \(L={\int }_{\alpha }^{\beta }\sqrt{{[f(\theta )]}^{2}+{[{f}^{'}(\theta )]}^{2}}d\theta ={\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{(\frac{dr}{d\theta })}^{2}}d\theta\) |
Area and Arc Length in Polar Coordinates
For the following exercises, determine a definite integral that represents the area.
For the following exercises, find the area of the described region.
For the following exercises, find a definite integral that represents the arc length.
For the following exercises, find the length of the curve over the given interval.
For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.
For the following exercises, use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.
For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.
Condensed: the full section is in OpenStax Calculus Volume 3.
Areas of Regions Bounded by Polar Curves
We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.
Consider a curve defined by the function \(r=f(\theta ),\) where \(\alpha \le \theta \le \beta .\) Our first step is to partition the interval \([\alpha ,\beta ]\) into n equal-width subintervals. The width of each subinterval is given by the formula \(\Delta \theta =(\beta -\alpha )\text{/}n,\) and the ith partition point \({\theta }_{i}\) is given by the formula \({\theta }_{i}=\alpha +i\Delta \theta .\) Each partition point \(\theta ={\theta }_{i}\) defines a line with slope \(\text{tan}{\theta }_{i}\) passing through the pole as shown in the following graph.
The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.
Recall that the area of a circle is \(A=\pi {r}^{2}.\) When measuring angles in radians, 360 degrees is equal to \(2\pi\) radians. Therefore a fraction of a circle can be measured by the central angle \(\theta .\) The fraction of the circle is given by \(\frac{\theta }{2\pi },\) so the area of the sector is this fraction multiplied by the total area:
\[A=(\frac{\theta }{2\pi })\ \pi {r}^{2}=\frac{1}{2}\theta {r}^{2}.\]Since the radius of a typical sector in is given by \({r}_{i}=f({\theta }_{i}),\) the area of the ith sector is given by
\[{A}_{i}=\frac{1}{2}(\Delta \theta ){(f({\theta }_{i}))}^{2}.\]Therefore a Riemann sum that approximates the area is given by
\[{A}_{n}=\sum _{i=1}^{n}{A}_{i}\approx \sum _{i=1}^{n}\frac{1}{2}(\Delta \theta ){(f({\theta }_{i}))}^{2}.\]We take the limit as \(n\to \infty\) to get the exact area:
\[A=\underset{n\to \infty }{\text{lim}}{A}_{n}=\frac{1}{2}{\int }_{\alpha }^{\beta }{(f(\theta ))}^{2}d\theta .\]\[6\ \text{sin}\ \theta =0.\]Condensed: the full section is in OpenStax Calculus Volume 2.
Arc Length in Polar Curves
Here we derive a formula for the arc length of a curve defined in polar coordinates.
In rectangular coordinates, the arc length of a parameterized curve \((x(t),y(t))\) for \(a\le t\le b\) is given by
\[L={\int }_{a}^{b}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt.\]In polar coordinates we define the curve by the equation \(r=f(\theta ),\) where \(\alpha \le \theta \le \beta .\) In order to adapt the arc length formula for a polar curve, we use the equations
\[x=r\ \text{cos}\ \theta =f(\theta )\ \text{cos}\ \theta \ \text{and}\ y=r\ \text{sin}\ \theta =f(\theta )\ \text{sin}\ \theta ,\]and we replace the parameter t by \(\theta .\) Then
\[\begin{array}{l}\frac{dx}{d\theta }={f}^{'}(\theta )\ \text{cos}\ \theta -f(\theta )\ \text{sin}\ \theta \\ \frac{dy}{d\theta }={f}^{'}(\theta )\ \text{sin}\ \theta +f(\theta )\ \text{cos}\ \theta .\end{array}\]We replace \(dt\) by \(d\theta ,\) and the lower and upper limits of integration are \(\alpha\) and \(\beta ,\) respectively. Then the arc length formula becomes
\[\begin{array}{ll}L & ={\int }_{a}^{b}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt \\ & ={\int }_{\alpha }^{\beta }\sqrt{{(\frac{dx}{d\theta })}^{2}+{(\frac{dy}{d\theta })}^{2}}d\theta \\ & ={\int }_{\alpha }^{\beta }\sqrt{{({f}^{'}(\theta )\ \text{cos}\ \theta -f(\theta )\ \text{sin}\ \theta )}^{2}+{({f}^{'}(\theta )\ \text{sin}\ \theta +f(\theta )\ \text{cos}\ \theta )}^{2}}d\theta \\ & ={\int }_{\alpha }^{\beta }\sqrt{{({f}^{'}(\theta ))}^{2}({\text{cos}}^{2}\theta +{\text{sin}}^{2}\theta )+{(f(\theta ))}^{2}({\text{cos}}^{2}\theta +{\text{sin}}^{2}\theta )}d\theta \\ & ={\int }_{\alpha }^{\beta }\sqrt{{({f}^{'}(\theta ))}^{2}+{(f(\theta ))}^{2}}d\theta \\ & ={\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{(\frac{dr}{d\theta })}^{2}}d\theta .\end{array}\]This gives us the following theorem.
Condensed: the full section is in OpenStax Calculus Volume 2.
Key Concepts
- The area of a region in polar coordinates defined by the equation \(r=f(\theta )\) with \(\alpha \le \theta \le \beta\) is given by the integral \(A=\frac{1}{2}{{\int }_{\alpha }^{\beta }[f(\theta )]}^{2}d\theta .\)
- To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
- The arc length of a polar curve defined by the equation \(r=f(\theta )\) with \(\alpha \le \theta \le \beta\) is given by the integral \(L={\int }_{\alpha }^{\beta }\sqrt{{[f(\theta )]}^{2}+{[{f}^{'}(\theta )]}^{2}}d\theta ={\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{(\frac{dr}{d\theta })}^{2}}d\theta .\)
Key Equations
| Area of a region bounded by a polar curve | \(A=\frac{1}{2}{\int }_{\alpha }^{\beta }{[f(\theta )]}^{2}d\theta =\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^{2}d\theta\) |
| Arc length of a polar curve | \(L={\int }_{\alpha }^{\beta }\sqrt{{[f(\theta )]}^{2}+{[{f}^{'}(\theta )]}^{2}}d\theta ={\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{(\frac{dr}{d\theta })}^{2}}d\theta\) |
Area and Arc Length in Polar Coordinates
For the following exercises, determine a definite integral that represents the area.
For the following exercises, find the area of the described region.
For the following exercises, find a definite integral that represents the arc length.
For the following exercises, find the length of the curve over the given interval.
For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.
For the following exercises, use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.
For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.
Condensed: the full section is in OpenStax Calculus Volume 2.
Sasa Hakuna kifaa cha kupigia hesabu kinachoweza kufanya hesabu, lakini kifaa hicho kinaweza kutumiwa katika hesabu.
Zoea (40)
Jaribu kila moja kwenye karatasi kwanza. kufunua jibu la kuchunguza; zile zilizothibitishwa zaweza kufunguliwa katika kitatua kwa kila hatua.
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Find the area of one petal of the rose defined by the equation \(r=3\ \text{sin}(2\theta ).\)
Lafunua jibu
The graph of \(r=3\ \text{sin}(2\theta )\) follows.
When \(\theta =0\) we have \(r=3\ \text{sin}(2(0))=0.\) The next value for which \(r=0\) is \(\theta =\pi \text{/}2.\) This can be seen by solving the equation \(3\ \text{sin}(2\theta )=0\) for \(\theta .\) Therefore the values \(\theta =0\) to \(\theta =\pi \text{/}2\) trace out the first petal of the rose. To find the area inside this petal, use with \(f(\theta )=3\ \text{sin}(2\theta ),\) \(\alpha =0,\) and \(\beta =\pi \text{/}2\text{:}\)
\[\begin{array}{ll}A & =\frac{1}{2}{\int }_{\alpha }^{\beta }{[f(\theta )]}^{2}d\theta \\ & =\frac{1}{2}{\int }_{0}^{\pi \text{/}2}{[3\ \text{sin}(2\theta )]}^{2}d\theta \\ & =\frac{1}{2}{\int }_{0}^{\pi \text{/}2}9\ {\text{sin}}^{2}(2\theta )\ d\theta .\end{array}\]To evaluate this integral, use the formula \({\text{sin}}^{2}\alpha =(1-\text{cos}(2\alpha ))\text{/}2\) with \(\alpha =2\theta \text{:}\)
\[\begin{array}{ll}A & =\frac{1}{2}{\int }_{0}^{\pi \text{/}2}9\ {\text{sin}}^{2}(2\theta )\ d\theta \\ & =\frac{9}{2}{\int }_{0}^{\pi \text{/}2}\frac{(1-\text{cos}(4\theta ))}{2}d\theta \\ & =\frac{9}{4}({\int }_{0}^{\pi \text{/}2}1-\text{cos}(4\theta )\ d\theta ) \\ & =\frac{9}{4}{(\theta -\frac{\text{sin}(4\theta )}{4})}_{0}^{\pi \text{/}2} \\ & =\frac{9}{4}(\frac{\pi }{2}-\frac{\text{sin}\ 2\pi }{4})-\frac{9}{4}(0-\frac{\text{sin}\ 4(0)}{4}) \\ & =\frac{9\pi }{8}.\end{array}\] -
Find the area inside the cardioid defined by the equation \(r=1-\text{cos}\ \theta .\)
Lafunua jibu
\(A=3\pi \text{/}2\)
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Find the area outside the cardioid \(r=2+2\ \text{sin}\ \theta\) and inside the circle \(r=6\ \text{sin}\ \theta .\)
Lafunua jibu
First draw a graph containing both curves as shown.
To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for \(\theta \text{:}\)
\[\begin{array}{lll}6\ \text{sin}\ \theta & = & 2+2\ \text{sin}\ \theta \\ 4\ \text{sin}\ \theta & = & 2 \\ \text{sin}\ \theta & = & \frac{1}{2}.\end{array}\]This gives the solutions \(\theta =\frac{\pi }{6}\) and \(\theta =\frac{5\pi }{6},\) which are the limits of integration. The circle \(r=3\ \text{sin}\ \theta\) is the red graph, which is the outer function, and the cardioid \(r=2+2\ \text{sin}\ \theta\) is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between \(\theta =\frac{\pi }{6}\) and \(\theta =\frac{5\pi }{6},\) then subtract the area inside the cardioid between \(\theta =\frac{\pi }{6}\) and \(\theta =\frac{5\pi }{6}\text{:}\)
\[\begin{array}{ll}A & =\text{circle}-\text{cardioid} \\ & =\frac{1}{2}{\int }_{\pi \text{/}6}^{5\pi \text{/}6}{[6\ \text{sin}\ \theta ]}^{2}d\theta -\frac{1}{2}{\int }_{\pi \text{/}6}^{5\pi \text{/}6}{[2+2\ \text{sin}\ \theta ]}^{2}d\theta \\ & =\frac{1}{2}{\int }_{\pi \text{/}6}^{5\pi \text{/}6}36\ {\text{sin}}^{2}\theta \ d\theta -\frac{1}{2}{\int }_{\pi \text{/}6}^{5\pi \text{/}6}[4+8\ \text{sin}\ \theta +4\ {\text{sin}}^{2}\theta ]\ d\theta \\ & =18{\int }_{\pi \text{/}6}^{5\pi \text{/}6}\frac{1-\text{cos}(2\theta )}{2}d\theta -2{\int }_{\pi \text{/}6}^{5\pi \text{/}6}[1+2\ \text{sin}\ \theta +\frac{1-\text{cos}(2\theta )}{2}]d\theta \\ & =9{[\theta -\frac{\text{sin}(2\theta )}{2}]}_{\pi \text{/}6}^{5\pi \text{/}6}-2{[\frac{3\theta }{2}-2\ \text{cos}\ \theta -\frac{\text{sin}(2\theta )}{4}]}_{\pi \text{/}6}^{5\pi \text{/}6} \\ & =9(\frac{5\pi }{6}-\frac{\text{sin}\ 2(5\pi \text{/}6)}{2})-9(\frac{\pi }{6}-\frac{\text{sin}\ 2(\pi \text{/}6)}{2}) \\ & \ \text{-}(3(\frac{5\pi }{6})-4\ \text{cos}\ \frac{5\pi }{6}-\frac{\text{sin}\ 2(5\pi \text{/}6)}{2})+(3(\frac{\pi }{6})-4\ \text{cos}\ \frac{\pi }{6}-\frac{\text{sin}\ 2(\pi \text{/}6)}{2}) \\ & =4\pi .\end{array}\] -
Find the area inside the circle \(r=4\ \text{cos}\ \theta\) and outside the circle \(r=2.\)
Lafunua jibu
\(A=\frac{4\pi }{3}+2\sqrt{3}\)
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Find the arc length of the cardioid \(r=2+2\ \text{cos}\ \theta .\)
Lafunua jibu
When \(\theta =0,r=2+2\ \text{cos}\ 0=4.\) Furthermore, as \(\theta\) goes from \(0\) to \(2\pi \text{,}\) the cardioid is traced out exactly once. Therefore these are the limits of integration. Using \(f(\theta )=2+2\ \text{cos}\ \theta ,\) \(\alpha =0,\) and \(\beta =2\pi \text{,}\) becomes
\[\begin{array}{ll}L & ={\int }_{\alpha }^{\beta }\sqrt{{[f(\theta )]}^{2}+{[{f}^{'}(\theta )]}^{2}}\ d\theta \\ & ={\int }_{0}^{2\pi }\sqrt{{[2+2\ \text{cos}\ \theta ]}^{2}+{[-2\ \text{sin}\ \theta ]}^{2}}\ d\theta \\ & ={\int }_{0}^{2\pi }\sqrt{4+8\ \text{cos}\ \theta +4\ {\text{cos}}^{2}\ \theta +4\ {\text{sin}}^{2}\ \theta }\ d\theta \\ & ={\int }_{0}^{2\pi }\sqrt{4+8\ \text{cos}\ \theta +4({\text{cos}}^{2}\ \theta +{\text{sin}}^{2}\ \theta )}\ d\theta \\ & ={\int }_{0}^{2\pi }\sqrt{8+8\ \text{cos}\ \theta }\ d\theta \\ & =2{\int }_{0}^{2\pi }\sqrt{2+2\ \text{cos}\ \theta }\ d\theta .\end{array}\]Next, using the identity \(\text{cos}(2\alpha )=2\ {\text{cos}}^{2}\alpha -1,\) add 1 to both sides and multiply by 2. This gives \(2+2\ \text{cos}(2\alpha )=4\ {\text{cos}}^{2}\alpha .\) Substituting \(\alpha =\theta \text{/}2\) gives \(2+2\ \text{cos}\ \theta =4\ {\text{cos}}^{2}(\theta \text{/}2),\) so the integral becomes
\[\begin{array}{ll}L & =2{\int }_{0}^{2\pi }\sqrt{2+2\ \text{cos}\ \theta }d\theta \\ & =2{\int }_{0}^{2\pi }\sqrt{4\ {\text{cos}}^{2}(\frac{\theta }{2})}d\theta \\ & =2{\int }_{0}^{2\pi }2|\text{cos}(\frac{\theta }{2})|d\theta .\end{array}\]The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue, change the limits from \(0\) to \(\pi\) and double the answer. This strategy works because cosine is positive between \(0\) and \(\frac{\pi }{2}.\) Thus,
\[\begin{array}{ll}L & =4{\int }_{0}^{2\pi }|\text{cos}(\frac{\theta }{2})|d\theta \\ & =8{\int }_{0}^{\pi }\text{cos}(\frac{\theta }{2})\ d\theta \\ & =8{(2\ \text{sin}(\frac{\theta }{2}))}_{0}^{\pi } \\ & =16.\end{array}\] -
Find the total arc length of \(r=3\ \text{sin}\ \theta .\)
Lafunua jibu
\(s=3\pi\)
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Region enclosed by \(r=4\)
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Region enclosed by \(r=3\ \text{sin}\ \theta\)
Lafunua jibu
\(\frac{9}{2}{\int }_{0}^{\pi }{\text{sin}}^{2}\theta \ d\theta\)
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Region in the first quadrant within the cardioid \(r=1+\text{sin}\ \theta\)
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Region enclosed by one petal of \(r=8\ \text{sin}(2\theta )\)
Lafunua jibu
\(32{\int }_{0}^{\pi \text{/}2}{\text{sin}}^{2}(2\theta )d\theta\)
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Region enclosed by one petal of \(r=\text{cos}(3\theta )\)
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Region below the polar axis and enclosed by \(r=1-\text{sin}\ \theta\)
Lafunua jibu
\(\frac{1}{2}{\int }_{\pi }^{2\pi }{(1-\text{sin}\ \theta )}^{2}d\theta\)
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Region in the first quadrant enclosed by \(r=2-\text{cos}\ \theta\)
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Region enclosed by the inner loop of \(r=2-3\ \text{sin}\ \theta\)
Lafunua jibu
\({\int }_{{\text{sin}}^{-1}(2\text{/}3)}^{\pi \text{/}2}{(2-3\ \text{sin}\ \theta )}^{2}d\theta\)
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Region enclosed by the inner loop of \(r=3-4\ \text{cos}\ \theta\)
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Region enclosed by \(r=1-2\ \text{cos}\ \theta\) and outside the inner loop
Lafunua jibu
\({\int }_{\pi \text{/}3}^{\pi }{(1-2\ \text{cos}\ \theta )}^{2}d\theta -{\int }_{0}^{\pi \text{/}3}{(1-2\ \text{cos}\ \theta )}^{2}d\theta\)
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Region common to \(r=3\ \text{sin}\ \theta \ \text{and}\ r=2-\text{sin}\ \theta\)
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Region common to \(r=2\ \text{and}\ r=4\ \text{cos}\ \theta\)
Lafunua jibu
\(4{\int }_{0}^{\pi \text{/}3}d\theta +16{\int }_{\pi \text{/}3}^{\pi \text{/}2}({\text{cos}}^{2}\theta )d\theta\)
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Region common to \(r=3\ \text{cos}\ \theta \ \text{and}\ r=3\ \text{sin}\ \theta\)
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Enclosed by \(r=6\ \text{sin}\ \theta\)
Lafunua jibu
\(9\pi\)
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Above the polar axis enclosed by \(r=2+\text{sin}\ \theta\)
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Below the polar axis and enclosed by \(r=2-\text{cos}\ \theta\)
Lafunua jibu
\(\frac{9\pi }{4}\)
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Enclosed by one petal of \(r=4\ \text{cos}(3\theta )\)
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Enclosed by one petal of \(r=3\ \text{cos}(2\theta )\)
Lafunua jibu
\(\frac{9\pi }{8}\)
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Enclosed by \(r=1+\text{sin}\ \theta\)
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Enclosed by the inner loop of \(r=3+6\ \text{cos}\ \theta\)
Lafunua jibu
\(\frac{18\pi -27\sqrt{3}}{2}\)
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Enclosed by \(r=2+4\ \text{cos}\ \theta\) and outside the inner loop
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Common interior of \(r=4\ \text{sin}(2\theta )\ \text{and}\ r=2\)
Lafunua jibu
\(\frac{4}{3}(4\pi -3\sqrt{3})\)
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Common interior of \(r=3-2\ \text{sin}\ \theta \ \text{and}\ r=-3+2\ \text{sin}\ \theta\)
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Common interior of \(r=6\ \text{sin}\ \theta \ \text{and}\ r=3\)
Lafunua jibu
\(\frac{3}{2}(4\pi -3\sqrt{3})\)
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Inside \(r=1+\text{cos}\ \theta\) and outside \(r=\text{cos}\ \theta\)
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Common interior of \(r=2+2\ \text{cos}\ \theta \ \text{and}\ r=2\ \text{sin}\ \theta\)
Lafunua jibu
\(2\pi -4\)
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\(r=4\ \text{cos}\ \theta \ \text{on the interval}\ 0\le \theta \le \frac{\pi }{2}\)
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\(r=1+\text{sin}\ \theta\) on the interval \(0\le \theta \le 2\pi\)
Lafunua jibu
\({\int }_{0}^{2\pi }\sqrt{{(1+\text{sin}\ \theta )}^{2}+{\text{cos}}^{2}\theta }d\theta\)
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\(r=2\ \text{sec}\ \theta \ \text{on the interval}\ 0\le \theta \le \frac{\pi }{3}\)
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\(r={e}^{\theta }\text{on the interval}\ 0\le \theta \le 1\)
Lafunua jibu
\(\sqrt{2}{\int }_{0}^{1}{e}^{\theta }d\theta\)
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\(r=6\ \text{on the interval}\ 0\le \theta \le \frac{\pi }{2}\)
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\(r={e}^{3\theta }\text{on the interval}\ 0\le \theta \le 2\)
Lafunua jibu
\(\frac{\sqrt{10}}{3}({e}^{6}-1)\)
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\(r=6\ \text{cos}\ \theta \ \text{on the interval}\ 0\le \theta \le \frac{\pi }{2}\)
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\(r=8+8\ \text{cos}\ \theta \ \text{on the interval}\ 0\le \theta \le \pi\)
Lafunua jibu
32
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Vaa alama yoyote ya ufafanuzi kamili, picha, na maana ya kila herufi.
Jinsi ya Kufaulu: Area and Arc Length in Polar Coordinates
- Apply the formula for area of a region in polar coordinates.
- Determine the arc length of a polar curve.
- The area of a region in polar coordinates defined by the equation
- To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
- The arc length of a polar curve defined by the equation
Maswali ambayo watu huuliza
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen: the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Sehemu za ukurasa huu zimebadilishwa kutoka kwa OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Tumekamatwa na kukosolewa tena hapa; makosa ni yetu. Kwa sababu chanzo chake kimeidhinishwa kwa matumizi yasiyo ya kibiashara, somo hili ni bure, haliuziwi chochote, na hushirikiwa na leseni ileile.
Mengi zaidi katika Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems