Skip to main content

Course Notes for Multivariable Calculus

Remkes Kooistra

Section 8.4 Week 8 Activity

Subsection 8.4.1 Polar Coordinates

Activity 8.4.1.

Use polar coordinates to solve this integral of the function \(f(x,y) = \sqrt{x^2 + y^2}\) on the wedge of the circle of radius \(3\) between \(\theta = \frac{3\pi}{4}\) and \(\theta = \frac{5\pi}{4}\text{.}\)
Solution.
Since \(r = \sqrt{x^2+y^2}\text{,}\) the integrand here is simply \(r\text{.}\) The region is a section of a circle, so the radius will vary from \(0\) to \(3\) and the angle between the two angles given. The Jacobian for polar coordinates is \(r\text{.}\)
\begin{gather*} \int_0^3 \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} r r d\theta dr = \frac{r^3}{3} \bigg|_0^3 \theta \bigg|_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} = 9 \left( \frac{5\pi}{4} - \frac{3\pi}{4} \right) = \frac{9\pi}{2} \end{gather*}

Activity 8.4.2.

Use polar coordinates to solve this integral of the function \(f(x,y) = x^2 - y^2\) on the quarter circle of radius \(4\) which is above line \(y=-x\text{,}\) below the line \(y=x\) and include the positive \(x\) axis.
Solution.
The region is a section of a circle. The radius will range from \(0\) to \(4\) and, for this section, I can take \(\theta \in \left[ \frac{-\pi}{4}, \frac{\pi}{4} \right]\text{.}\) For the integrand, I replace \(x = r \cos \theta\) and \(y = r \sin \theta\) to get \(r^2(\cos^2 \theta - \sin^2 \theta)\text{.}\) Using trig identities, I can write this as \(r^2 \cos 2\theta\text{.}\) The Jacobian of polar coordinates is \(r\text{.}\)
\begin{align*} \int_0^4 \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} r^2 \cos (2\theta) r d\theta dr \amp = \int_0^4 r^3 dr \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \cos (2\theta) d \theta = \frac{r^4}{4} \bigg|_0^4 \frac{\sin 2\theta}{2} \bigg|_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \\ \amp = 32 \left( \sin \left( \frac{\pi}{2} \right) - \sin \left( \frac{-\pi}{2} \right) \right) = 32(2) = 64 \end{align*}

Activity 8.4.3.

Use polar coordinates to solve this integral of the function \(f(x,y) = (x^2 + y^2)^{\frac{3}{2}}\) on the unit circle centred at \((0,1)\text{.}\)
Solution.
The region is a an offset circle, so I need its in polar locus. In cartesian coordinates, the circle is \(x^2 + (y-1)^2 = 1\text{.}\) If I replace the cartesian coordinates with their polar replacements and simplifity, I get these calculations.
\begin{align*} r^2 \cos^2 \theta + (r \sin \theta - 1)^2 \amp = 1 \\ r^2 \cos^2 \theta + r^2 \sin^2 \theta - 2r \sin \theta + 1 \amp = 1 \\ r^2 - 2r \sin \theta \amp = 0 \\ r = 2 \sin \theta \end{align*}
I can take \(r \in [0, 2 \sin \theta]\text{,}\) since the circle touches the origin at one point. The circle is in the half-plane where \(y\) is positive, so the appropriate range of the angle is \(\theta \in [0, \pi]\text{.}\) If I use \(\theta\) as the outside integral, this gives me a way to setup an interated integral with non-constant bounds in polar coordinates. The integral simplifies to \(r^3\text{.}\) (A asked a computer for the antiderivative of \(\sin^5 \theta\text{.}\))
\begin{align*} \int_0^{pi} \int_0^{2 \sin} \theta r^3 rdr d\theta \amp = \int_0^{\pi} \frac{r^5}{5} \Bigg|_{0}^{2\sin theta} d\theta \\ \amp = \frac{32}{5} \int_0^{\pi} \sin^5 \theta d \theta \\ \amp = \frac{32}{5} \left( \frac{-5 \cos x}{8} + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos(5x) \right) \Bigg|_0^{\pi} \\ \amp = \frac{32}{5} \left( \frac{-5 \cos \pi}{8} + \frac{5}{48} \cos (3\pi) - \frac{1}{80} \cos (5\pi) \right) \\ \amp - \left( \frac{-5 \cos 0}{8} + \frac{5}{48} \cos (0) - \frac{1}{80} \cos(0) \right) \\ \amp = \frac{32}{5} \left( \left( \frac{5}{8} - \frac{5}{48} + \frac{1}{80} \right) - \left( \frac{5}{8} + \frac{5}{48} - \frac{1}{80} \right) \right) \\ \amp = \frac{32}{5} \left( \frac{10}{8} - \frac{10}{48} + \frac{2}{80} \right) = \frac{712}{75} \end{align*}

Activity 8.4.4.

Use polar coordinates to solve this integral of the function \(f(x,y) = 3x^3 - (y-3)^2\) on circle of radius \(2\) centred at the origin.
Solution.
The region here is easy: \(r \in [0,2]\) and \(\theta \in [0, 2\pi]\text{.}\) The integrand, however, doesn’t look that pleasant after I change to polar coordinates: \(3r^3 \cos^3 \theta - (r \sin \theta - 3)^2\text{.}\) Adding the Jacobian gives the new integral. I split it up into several pieces to complete the integral. I did some of the trig integrals using a computer algebra system (though you can see, directly, that some evaluate to zero using symmetry arguments.)
\begin{align*} \amp \int_0^{2\pi} \int_0^2 (3r^3 \cos^3 \theta - r^2 \sin^2 \theta + 6r \sin \theta - 9) r dr d\theta \\ \amp = \int_0^{2\pi} \int_0^2 3r^4 \cos^3 \theta dr d\theta - \int_0^{2\pi} \int_0^2 r^3 \sin^2 \theta dr d\theta \\ \amp + \int_0^{2\pi} \int_0^2 6r^2 \sin \theta dr d\theta - \int_0^{2\pi} \int_0^2 9r dr d\theta \\ \amp = \int_0^{2\pi} \cos^3 \theta d\theta \int_0^2 3r^4 dr - \int_0^{2\pi} \sin^2 \theta d\theta \int_0^2 r^3 dr\\ \amp + \int_0^{2\pi} \sin \theta d\theta \int_0^2 6r^2 dr - \int_0^{2\pi} d\theta \int_0^2 9r dr \\ \amp = (0) \frac{3r^5}{5} \bigg|_0^2 - (\pi) \frac{r^4}{4} \bigg|_0^2 + (0) 2r^3 \bigg|_0^2 - (2\pi) \frac{9r^2}{2} \bigg|_0^2 \\ \amp = 0 - 4\pi + 0 - 36\pi = -40\pi \end{align*}

Subsection 8.4.2 Spherical and Cylindrical Coordinates

Activity 8.4.5.

Integrate \(f(x,y,z) = 2x^2 + y^2 +z^2\) on the sphere of radius \(2\) centred at the origin.
Solution.
The sphere is described nicely in spherical coordinates as \(r \in [0,2]\text{,}\) \(\theta \in [0, 2\pi]\) and \(\phi \in [0, \pi]\text{.}\) The integrand can be seperated as \(x^2 + (x^2 + y^2 + z^2)\) and written as \(r^2 \sin^2 \phi \cos^2 \theta + r^2\text{.}\) I can write the integral in spherical coordinates and split it up into two integrals. The Jacobian is \(r^2 \sin \phi\text{.}\)
\begin{align*} \amp \int_0^2 \int_0^{2\pi} \int_0^{\pi} (r^2 \sin^2 \phi \cos^2 \theta + r^2) r^2 \sin \phi d\phi d\theta dr \\ \amp = \int_0^2 \int_0^{2\pi} \int_0^{\pi} r^3 \sin^3 \phi \cos^2 \theta d\phi d\theta dr + \int_0^2 \int_0^{2\pi} \int_0^{\pi} r^4 \sin \phi d\phi d\theta dr \\ \amp = \int_0^2 r^4 dr \int_0^{2\pi} \cos^2 \theta d \theta \int_0^{\pi} \sin^3 \phi d\phi + \int_0^2 r^4 dr \int_0^{2\pi} d\theta \int_0^{\pi} \sin \phi d\phi \\ \amp = \left( \frac{32}{5} \right) (\pi)\left( \frac{4}{3} \right) + \left( \frac{32}{5} \right) (2\pi) (2) = \frac{128\pi}{15} + \frac{128\pi}{5} = \frac{512 \pi}{15} \end{align*}

Activity 8.4.6.

Integrate \(f(x,y,z) = x^2 + y^2 - 4z\) on the cylinder of radius \(4\) about the \(z\) axis restricted to the height \(z \in [-1,3]\text{.}\)
Solution.
The cylinder is described nicely in cylindrical coordinates as \(r \in [0,4]\text{,}\) \(\theta \in [0, 2\pi]\) and \(z \in [-1,3]\text{.}\) The integrand becomes \(r^2 - 4z\text{.}\) The Jacobian is \(r\text{.}\)
\begin{align*} \amp \int_0^4 \int_0^{2\pi} \int_{-1}^3 (r^2 - 4z) r dz d\theta dr = \int_0^4 \int_0^{2\pi} \int_{-1}^3 r^3 dz d\theta dr - \int_0^4 \int_0^{2\pi} \int_{-1}^3 4rz dz d\theta dr \\ \amp = \int_0^4 r^3 dr \int_0^{2\pi} d\theta \int_{-1}^3 dz - 4 \int_0^4 r dr \int_0^{2\pi} d\theta \int_{-1}^3 z dz \\ \amp = (16)(2\pi)(4) - 4 (8)(2\pi)(4) = 128 \pi - 256 \pi = 128 \pi \end{align*}

Activity 8.4.7.

Integrate \(f(x,y,z) = x^2 + y^2\) on he sphere of radius \(1\) centred at the \((0,0,1\text{.}\)
Solution.
Even though the region is a sphere, the integrand looks better in cylindrical coordinates, so I’ll try that. The integrand simply becomes \(r^2\text{.}\) The sphere has cartesian equation \(x^2 + y^2 + (z-1)^2 = 1\text{.}\) In cylindrical coordinates, this becomes \(r^2 + (z-1)^2 = 1\text{.}\) I can take \(z \in [0,2]\) and \(r\) ranging from \(0\) to \(r = \sqrt{1-(z-1)^2} = \sqrt{2z-z^2}\text{.}\) The Jacobian is \(r\text{.}\)
\begin{align*} \amp \int_0^{2\pi} \int_0^2 \int_0^{\sqrt{2z-z^2}} r^2 r dr dz d\theta = \int_0^{2\pi} d\theta \int_0^2 \int_0^{\sqrt{2z-z^2}} r^3 dr dz \\ \amp = (2\pi) \int_0^2 \frac{r^4}{4} \Bigg|_0^{\sqrt{2z-z^2}} dz = \frac{\pi}{2} \int_0^2 (2z-z^2)^2 dz \\ \amp = \frac{\pi}{2} \int_0^2 z^4 - 4z^3 + 4z^2 dz = \frac{\pi}{2} \left( \frac{z^5}{5} - z^4 + \frac{4z^3}{3} \right) \Bigg|_0^2 \\ \amp = \frac{\pi}{2} \left( \frac{32}{5} - 16 + \frac{64}{3} \right) = \frac{88\pi}{15} \end{align*}

Activity 8.4.8.

Find the volume of the sphere of radius \(3\) with the parabaloid \(z = (x^2 + y^2) - 3\) cut out of it.
Solution.
Though I am dealing with a sphere, the parabaloid is easier to describe in cylindrical coordinates, so I will use that system. The whole system is symetrical about the \(z\) axis, so \(\theta\) can simply range from \(0\) to \(2\pi\text{.}\) The sphere is the outside range of \(r\) and the paraboloid is the inside range of \(r\text{,}\) but both depend on \(z\text{.}\) I need equations for both. The sphere is \(x^2 + y^2 +z^2 = 9\text{,}\) which changes to \(r^2 + z^2 = 9\text{.}\) Solving for \(r\) gives \(r = \sqrt{9-z^2}\text{.}\) The parabaloid is \(z = r^2 - 3\text{,}\) which simplifies to \(r = \sqrt{z+3}\text{.}\) The range on \(z\) starts at \(-3\) and ends at the point where the paraboloid and the sphere meet. In the two equations, I can isolate the replace \(r^2\) to get \(z+3+z^2 = 9\text{,}\) which is the quadratic \(z^2 + z - 6 = (z-3)(z+2)\text{.}\) Therefore, the parabaloid and sphere meet at the \(z=2\) plane and \(z=2\) is the upper bound for \(z\text{.}\) The Jacobian is \(r\text{.}\)
\begin{align*} \amp \int_0^{2\pi} \int_{-3}^2 \int_{\sqrt{z+3}}^{\sqrt{9-z^2}} r dr dz d\theta \\ \amp = \int_0^{2\pi} d\theta \int_{-3}^2 \frac{r^2}{2} \Bigg|_{\sqrt{z+3}}^{\sqrt{9-z^2}} dz \\ \amp = 2\pi \int_{-3}^2 \frac{1}{2} \left( 9-z^2 - (z+3) \right) dz\\ \amp = \pi \int_{-3}^2 6 - z - z^2 dz \\ \amp = \pi \left( 6z - \frac{z^2}{2} - \frac{z^3}{3} \right) \Bigg|_{-3}^2\\ \amp = \pi \left( 6(2-(-3)) - \frac{4 - 9}{2} - \frac{8 - (-27)}{3} \right) = \pi \left(30 + \frac{5}{2} - \frac{35}{3} \right) = \frac{125\pi}{6} \end{align*}

Activity 8.4.9.

Find the volume of the portion of the cylinder about the \(z\) axis of radius \(3\) which is above the \(z = -1\) plane but below the \(x + y + z = 5\) plane.
Solution.
I can work in cylindrical coordinates. The planes \(z = -1\) remains \(z=-1\text{.}\) The plane \(x + y + z = 5\) is trickier. In changed into \(r \cos \theta + r \sin \theta + z = 5 = r (\cos \theta + \sin \theta) + z = 5\text{.}\) I can use a trig identity to write this as \(r \left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right) + z = 5\text{.}\) This is tricky, since the equation involves all of the variables. I choose to set \(z\) first, then \(\theta\text{,}\) then \(r\text{.}\) For \(z\text{,}\) I need to calculate the maximum value; the highest point where the cylinder meets the plane. This happens when \(x = y = -3\text{,}\) so \(z=11\) is the highest value and I can take \(z \in [-1,11]\text{.}\) Then at each value of \(z\text{,}\) I have the relationship \(r \left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right) + z = 5\) between \(r\) and \(\theta\text{.}\) If I solve for \(r\text{,}\) I get
\begin{equation*} r = \frac{5-z}{\left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right)} \end{equation*}
This gets particularly tricky. I can’t choose either \(r\) or \(\theta\) to be the outside variable for the whole cylinder; I need two different cases for \(z \lt 5\) and \(z \gt 5\text{.}\) On the bottom half of the cylinder, I can use the previous equation to bound \(r\) for \(\theta \in \left[\frac{-\pi}{4} \frac{3\pi}{4} \right]\text{,}\) but \(r\) goes from \(0\) to \(3\) for the rest of the range. For the tope half of the cylinder, the bound of \(r\) in terms of \(\theta\) works for \(\theta in \left[ \frac{3\pi}{4}, \frac{7\pi}{4} \right]\text{.}\) For the rest of the range of \(\theta\text{,}\) there isn’t anything at all. This gives three integrals.
\begin{align*} V \amp = \int_{-1}^5 \int_{\frac{-\pi}{4}}^{\frac{3\pi}{4}} \int_0^{\frac{5-z}{\left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right)}} r dr d\theta dz + \int_{-1}^5 \int_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} \int_0^3 r dr d\theta dz\\ \amp \hspace{2cm} + \int_5^{11}\int_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} \int_0^{\frac{5-z}{\left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right)} }r dr d\theta dz \end{align*}
This is a miserable set of integrals, to be sure. Luckly for us, there is another way to approach this entire question. If we look at the full cylinder for \(z \in [-1,11]\text{,}\) the plane cuts this cylinder into two equal pieces. Therefore, the desired volume should be one half of the volume of the cylinder of height \(12\) and radius \(3\text{,}\) which is \(72\pi\text{.}\)

Subsection 8.4.3 Conceptual Review Questions

  • What are polar coordinates?
  • How is it that change of variable can be arranged to help either the integrand or the region (and sometimes both)?
  • What is a Jacobian and what does it measure?
  • What are the advantages of curvilinear coordinate systems?
<Prev^TopNext>