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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.

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  1. Find the area of one petal of the rose defined by the equation \(r=3\ \text{sin}(2\theta ).\)

    答えを表示

    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 \(\alpha =2\theta \text{:}\)

    \[\begin{array}{ll}A & =\frac{1}{2}{\int }_{0}^{\pi \text{/}2}9\ {\text{sin}}^{2}(2\theta )\ d\theta \\ & =\frac{9}{2}{\int }_{0}^{\pi \text{/}2}\frac{(1-\text{cos}(4\theta ))}{2}d\theta \\ & =\frac{9}{4}({\int }_{0}^{\pi \text{/}2}1-\text{cos}(4\theta )\ d\theta ) \\ & =\frac{9}{4}{(\theta -\frac{\text{sin}(4\theta )}{4})}_{0}^{\pi \text{/}2} \\ & =\frac{9}{4}(\frac{\pi }{2}-\frac{\text{sin}\ 2\pi }{4})-\frac{9}{4}(0-\frac{\text{sin}\ 4(0)}{4}) \\ & =\frac{9\pi }{8}.\end{array}\]

  2. Find the area inside the cardioid defined by the equation \(r=1-\text{cos}\ \theta .\)

    答えを表示

    \(A=3\pi \text{/}2\)

  3. Find the area outside the cardioid \(r=2+2\ \text{sin}\ \theta\) and inside the circle \(r=6\ \text{sin}\ \theta .\)

    答えを表示

    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 \(\theta =\frac{5\pi }{6}\text{:}\)

    \[\begin{array}{ll}A & =\text{circle}-\text{cardioid} \\ & =\frac{1}{2}{\int }_{\pi \text{/}6}^{5\pi \text{/}6}{[6\ \text{sin}\ \theta ]}^{2}d\theta -\frac{1}{2}{\int }_{\pi \text{/}6}^{5\pi \text{/}6}{[2+2\ \text{sin}\ \theta ]}^{2}d\theta \\ & =\frac{1}{2}{\int }_{\pi \text{/}6}^{5\pi \text{/}6}36\ {\text{sin}}^{2}\theta \ d\theta -\frac{1}{2}{\int }_{\pi \text{/}6}^{5\pi \text{/}6}[4+8\ \text{sin}\ \theta +4\ {\text{sin}}^{2}\theta ]\ d\theta \\ & =18{\int }_{\pi \text{/}6}^{5\pi \text{/}6}\frac{1-\text{cos}(2\theta )}{2}d\theta -2{\int }_{\pi \text{/}6}^{5\pi \text{/}6}[1+2\ \text{sin}\ \theta +\frac{1-\text{cos}(2\theta )}{2}]d\theta \\ & =9{[\theta -\frac{\text{sin}(2\theta )}{2}]}_{\pi \text{/}6}^{5\pi \text{/}6}-2{[\frac{3\theta }{2}-2\ \text{cos}\ \theta -\frac{\text{sin}(2\theta )}{4}]}_{\pi \text{/}6}^{5\pi \text{/}6} \\ & =9(\frac{5\pi }{6}-\frac{\text{sin}\ 2(5\pi \text{/}6)}{2})-9(\frac{\pi }{6}-\frac{\text{sin}\ 2(\pi \text{/}6)}{2}) \\ & \ \text{-}(3(\frac{5\pi }{6})-4\ \text{cos}\ \frac{5\pi }{6}-\frac{\text{sin}\ 2(5\pi \text{/}6)}{2})+(3(\frac{\pi }{6})-4\ \text{cos}\ \frac{\pi }{6}-\frac{\text{sin}\ 2(\pi \text{/}6)}{2}) \\ & =4\pi .\end{array}\]

  4. Find the area inside the circle \(r=4\ \text{cos}\ \theta\) and outside the circle \(r=2.\)

    答えを表示

    \(A=\frac{4\pi }{3}+2\sqrt{3}\)

  5. Find the arc length of the cardioid \(r=2+2\ \text{cos}\ \theta .\)

    答えを表示

    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,

    \[\begin{array}{ll}L & =4{\int }_{0}^{2\pi }|\text{cos}(\frac{\theta }{2})|d\theta \\ & =8{\int }_{0}^{\pi }\text{cos}(\frac{\theta }{2})\ d\theta \\ & =8{(2\ \text{sin}(\frac{\theta }{2}))}_{0}^{\pi } \\ & =16.\end{array}\]

  6. Find the total arc length of \(r=3\ \text{sin}\ \theta .\)

    答えを表示

    \(s=3\pi\)

  7. Region enclosed by \(r=4\)

  8. Region enclosed by \(r=3\ \text{sin}\ \theta\)

    答えを表示

    \(\frac{9}{2}{\int }_{0}^{\pi }{\text{sin}}^{2}\theta \ d\theta\)

  9. Region in the first quadrant within the cardioid \(r=1+\text{sin}\ \theta\)

  10. Region enclosed by one petal of \(r=8\ \text{sin}(2\theta )\)

    答えを表示

    \(32{\int }_{0}^{\pi \text{/}2}{\text{sin}}^{2}(2\theta )d\theta\)

  11. Region enclosed by one petal of \(r=\text{cos}(3\theta )\)

  12. Region below the polar axis and enclosed by \(r=1-\text{sin}\ \theta\)

    答えを表示

    \(\frac{1}{2}{\int }_{\pi }^{2\pi }{(1-\text{sin}\ \theta )}^{2}d\theta\)

  13. Region in the first quadrant enclosed by \(r=2-\text{cos}\ \theta\)

  14. Region enclosed by the inner loop of \(r=2-3\ \text{sin}\ \theta\)

    答えを表示

    \({\int }_{{\text{sin}}^{-1}(2\text{/}3)}^{\pi \text{/}2}{(2-3\ \text{sin}\ \theta )}^{2}d\theta\)

  15. Region enclosed by the inner loop of \(r=3-4\ \text{cos}\ \theta\)

  16. Region enclosed by \(r=1-2\ \text{cos}\ \theta\) and outside the inner loop

    答えを表示

    \({\int }_{\pi \text{/}3}^{\pi }{(1-2\ \text{cos}\ \theta )}^{2}d\theta -{\int }_{0}^{\pi \text{/}3}{(1-2\ \text{cos}\ \theta )}^{2}d\theta\)

  17. Region common to \(r=3\ \text{sin}\ \theta \ \text{and}\ r=2-\text{sin}\ \theta\)

  18. Region common to \(r=2\ \text{and}\ r=4\ \text{cos}\ \theta\)

    答えを表示

    \(4{\int }_{0}^{\pi \text{/}3}d\theta +16{\int }_{\pi \text{/}3}^{\pi \text{/}2}({\text{cos}}^{2}\theta )d\theta\)

  19. Region common to \(r=3\ \text{cos}\ \theta \ \text{and}\ r=3\ \text{sin}\ \theta\)

  20. Enclosed by \(r=6\ \text{sin}\ \theta\)

    答えを表示

    \(9\pi\)

  21. Above the polar axis enclosed by \(r=2+\text{sin}\ \theta\)

  22. Below the polar axis and enclosed by \(r=2-\text{cos}\ \theta\)

    答えを表示

    \(\frac{9\pi }{4}\)

  23. Enclosed by one petal of \(r=4\ \text{cos}(3\theta )\)

  24. Enclosed by one petal of \(r=3\ \text{cos}(2\theta )\)

    答えを表示

    \(\frac{9\pi }{8}\)

  25. Enclosed by \(r=1+\text{sin}\ \theta\)

  26. Enclosed by the inner loop of \(r=3+6\ \text{cos}\ \theta\)

    答えを表示

    \(\frac{18\pi -27\sqrt{3}}{2}\)

  27. Enclosed by \(r=2+4\ \text{cos}\ \theta\) and outside the inner loop

  28. Common interior of \(r=4\ \text{sin}(2\theta )\ \text{and}\ r=2\)

    答えを表示

    \(\frac{4}{3}(4\pi -3\sqrt{3})\)

  29. Common interior of \(r=3-2\ \text{sin}\ \theta \ \text{and}\ r=-3+2\ \text{sin}\ \theta\)

  30. Common interior of \(r=6\ \text{sin}\ \theta \ \text{and}\ r=3\)

    答えを表示

    \(\frac{3}{2}(4\pi -3\sqrt{3})\)

  31. Inside \(r=1+\text{cos}\ \theta\) and outside \(r=\text{cos}\ \theta\)

  32. Common interior of \(r=2+2\ \text{cos}\ \theta \ \text{and}\ r=2\ \text{sin}\ \theta\)

    答えを表示

    \(2\pi -4\)

  33. \(r=4\ \text{cos}\ \theta \ \text{on the interval}\ 0\le \theta \le \frac{\pi }{2}\)

  34. \(r=1+\text{sin}\ \theta\) on the interval \(0\le \theta \le 2\pi\)

    答えを表示

    \({\int }_{0}^{2\pi }\sqrt{{(1+\text{sin}\ \theta )}^{2}+{\text{cos}}^{2}\theta }d\theta\)

  35. \(r=2\ \text{sec}\ \theta \ \text{on the interval}\ 0\le \theta \le \frac{\pi }{3}\)

  36. \(r={e}^{\theta }\text{on the interval}\ 0\le \theta \le 1\)

    答えを表示

    \(\sqrt{2}{\int }_{0}^{1}{e}^{\theta }d\theta\)

  37. \(r=6\ \text{on the interval}\ 0\le \theta \le \frac{\pi }{2}\)

  38. \(r={e}^{3\theta }\text{on the interval}\ 0\le \theta \le 2\)

    答えを表示

    \(\frac{\sqrt{10}}{3}({e}^{6}-1)\)

  39. \(r=6\ \text{cos}\ \theta \ \text{on the interval}\ 0\le \theta \le \frac{\pi }{2}\)

  40. \(r=8+8\ \text{cos}\ \theta \ \text{on the interval}\ 0\le \theta \le \pi\)

    答えを表示

    32

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どうやって: Area and Arc Length in Polar Coordinates

  1. Apply the formula for area of a region in polar coordinates.
  2. Determine the arc length of a polar curve.
  3. The area of a region in polar coordinates defined by the equation
  4. To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
  5. The arc length of a polar curve defined by the equation

よくある質問

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.

Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). ここでは要約して説明し直しています。誤りは私たちの責任です。 ソースは非商用利用のためにライセンスされているため、このレッスンは無料で、売ることはなく、同じライセンスの下で共有されている。

More in Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems