Coordinated Calculus

First, we observe that \(y = 4-x^2\) intersects the \(x\)-axis at the points \((-2,0)\) and \((2,0)\text{.}\) When we revolve the region \(R\) about the \(x\)-axis, we get the three-dimensional solid pictured in Figure6.19.

We slice the solid into vertical slices of thickness \(\Delta x\) between \(x = -2\) and \(x = 2\text{.}\) A representative slice is a cylinder of height \(\Delta x\) and radius \(4-x^2\text{.}\) Hence, the volume of the slice is

Using a definite integral to sum the volumes of the representative slices, it follows that

It is straightforward to evaluate the integral and find that the volume is \(V = \frac{512}{15}\pi\text{.}\)

First, we must determine where the curves \(y = 4-x^2\) and \(y = x+2\) intersect. Substituting the expression for \(y\) from the second equation into the first equation, we find that \(x + 2 = 4-x^2\text{.}\) Rearranging, it follows that

and the solutions to this equation are \(x = -2\) and \(x = 1\text{.}\) The curves therefore cross at \((-2,0)\) and \((1,1)\text{.}\)

When we revolve the region \(R\) about the \(x\)-axis, we get the three-dimensional solid pictured at left in Figure6.21.

Immediately we see a major difference between the solid in this example and the one in Example6.18: here, the three-dimensional solid of revolution isn't solid because it has open space in its center along the axis of revolution. If we slice the solid perpendicular to the axis of revolution, we observe that the resulting slice is not a solid disk, but rather a washer, as pictured at right in Figure6.21. At a given location \(x\) between \(x = -2\) and \(x = 1\text{,}\) the small radius \(r(x)\) of the inner circle is determined by the curve \(y = x+2\text{,}\) so \(r(x) = x+2\text{.}\) Similarly, the big radius \(R(x)\) comes from the function \(y = 4-x^2\text{,}\) and thus \(R(x) = 4-x^2\text{.}\)

To find the volume of a representative slice, we compute the volume of the outer disk and subtract the volume of the inner disk. Since

it follows that the volume of a typical slice is

Using a definite integral to sum the volumes of the respective slices across the integral, we find that

Evaluating the integral, we find that the volume of the solid of revolution is \(V = \frac{108}{5}\pi\text{.}\)

1. Use slices perpendicular to the \(x\)-axis.

2. Note that slices perpendicular to the \(x\)-axis will be washers.

3. Find the points where the two curves intersect and draw a labeled graph.

4. Think about how the solid looks like a donut; what is the outer radius?

5. Try using slices perpendicular to the \(y\)-axis.

1. \(V = \int_0^4 \pi (\sqrt{x})^2 \, dx = \int_0^4 \pi x \, dx = 8\pi\text{.}\)

2. \(V = \int_0^4 \pi (4-(\sqrt{x})^2) \, dx = \int_0^4 \pi (4-x) \, dx = 8\pi\text{.}\)

3. \(V = \int_0^1 \pi(x - x^6) \, dx = \frac{5}{14}\pi\text{.}\)

4. \(V = \int_{-\sqrt{3}}^{\sqrt{3}} \pi( (x^2 + 4)^2 - (2x^2 + 1)^2) \, dx = \frac{136\sqrt{3}}{5}\pi\text{.}\)

5. \(V = \int_0^2 \pi y^4 \, dy = \frac{32}{5}\pi\text{.}\)

1. A typical slice (perpendicular to the \(x\)-axis) has radius \(r(x) = \sqrt{x}\text{,}\) and thus the volume of such a slice is

   \begin{equation*} V_{\text{slice}} = \pi (\sqrt{x})^2 \Delta x\text{.} \end{equation*}

   It follows by the disk method that the volume of the solid of revolution is

   \begin{equation*} V = \int_0^4 \pi (\sqrt{x})^2 \, dx = \int_0^4 \pi x \, dx = 8\pi\text{.} \end{equation*}

2. A typical slice (perpendicular to the \(x\)-axis) is a washer with outer radius \(R(x) = 2\) and inner radius \(r(x) = \sqrt{x}\text{.}\) The volume of such a slice is

   \begin{equation*} V_{\text{slice}} = \pi(R(x)^2-r(x)^2) \Delta x = \pi (2^2 - (\sqrt{x})^2) \Delta x\text{.} \end{equation*}

   It follows by the washer method that the volume of the solid of revolution is

   \begin{equation*} V = \int_0^4 \pi (4-(\sqrt{x})^2) \, dx = \int_0^4 \pi (4-x) \, dx = 8\pi\text{.} \end{equation*}

3. The curves \(y = \sqrt{x}\) and \(y = x^3\) intersect at \((0,0)\) and \((1,1)\text{;}\) revolving the region \(S\) about the \(x\)-axis and taking slices perpendicular to the \(x\)-axis, we see that a typical slice is a washer with outer radius \(R(x) = \sqrt{x}\) and inner radius \(r(x) = x^3\text{.}\) Hence the volume of a typical slice is

   \begin{equation*} V_{\text{slice}} = \pi(R(x)^2-r(x)^2) \Delta x = \pi((\sqrt{x})^2 - (x^3)^2) \Delta x = \pi(x - x^6) \Delta x\text{.} \end{equation*}

   By the washer method, the volume of the solid is

   \begin{equation*} V = \int_0^1 \pi(x - x^6) \, dx = \frac{5}{14}\pi\text{.} \end{equation*}

4. The curves \(y = 2x^2 + 1\) and \(y = x^2 + 4\) intersect at the points \((\pm \sqrt{3}, 7)\text{;}\) after we revolve the region \(S\) about the \(x\)-axis, cross-sections perpendicular to the axis of rotation are washers with outer radius \(R(x) = x^2 + 4\) and inner radius \(r(x) = 2x^2 + 1\text{.}\) By the washer method, we see that

   \begin{equation*} V = \int_{-\sqrt{3}}^{\sqrt{3}} \pi( (x^2 + 4)^2 - (2x^2 + 1)^2) \, dx = \frac{136\sqrt{3}}{5}\pi\text{.} \end{equation*}

5. The curve \(y = \sqrt{x}\) and the line \(y = 2\) intersect at \((4,2)\text{;}\) when we revolve the region \(S\) about the \(y\)-axis, slices perpendicular to the \(y\)-axis are disks. Here we see that each disk's radius is a function of \(y\) and the disks have thickness \(\Delta y\text{,}\) so we will need to integrate with respect to \(y\) and consider the radius function \(r(y) = y^2\) (which comes from solving \(y = \sqrt{x}\) for \(x\) in terms of \(y\text{.}\) Note how this is different from the problem in part (b) because of the axis we rotate about and how this forces us to integrate in terms of \(y\text{.}\) A typical slice has volume

   \begin{equation*} V_{\text{slice}} = \pi (y^2)^2 \Delta y\text{,} \end{equation*}

   and thus by the disk method the total volume is

   \begin{equation*} V = \int_0^2 \pi y^4 \, dy = \frac{32}{5}\pi\text{.} \end{equation*}

These two curves intersect when \(x = 1\text{,}\) hence at the point \((1,1)\text{.}\) When we revolve the region \(R\) about the \(y\)-axis, we get the three-dimensional solid pictured at left in Figure6.24.

Note that the slices are cylindrical washers only if taken perpendicular to the \(y\)-axis. We slice the solid horizontally, starting at \(y = 0\) and proceeding up to \(y = 1\text{.}\) The thickness of a representative slice is \(\Delta y\text{,}\) so we must express the integrand in terms of \(y\text{.}\) The inner radius is determined by the curve \(y = \sqrt{x}\text{,}\) so we solve for \(x\) and get \(x = y^2 = r(y)\text{.}\) In the same way, we solve the curve \(y = x^4\) (which governs the outer radius) for \(x\) in terms of \(y\text{,}\) and hence \(x = \sqrt[4]{y}\text{.}\) Therefore, the volume of a typical slice is

We use a definite integral to sum the volumes of all the slices from \(y = 0\) to \(y = 1\text{.}\) The total volume is

It is straightforward to evaluate the integral and find that \(V = \frac{7}{15} \pi\text{.}\)

1. Consider slices perpendicular to the \(y\)-axis.

2. Since we revolve around the \(y\)-axis, think about slices of thickness \(\Delta y\text{.}\)

3. Note that \(y = x^3\) lies below \(y = 2x\) on the interval \([0,\sqrt{2}]\text{.}\)

4. The curve \(y = x^3\) can be equivalently expressed as \(x = \sqrt[3]{y}\text{.}\)

5. Find the intersection points by writing the second curve as \(x = y+1\) and solving \((y-1)^2 = y+1\text{.}\)

1. \(V = \int_0^2 \pi y^4 \Delta \, dy\text{.}\)

2. \(V = \int_0^2 \pi (16 - y^4) \, dy\text{.}\)

3. \(V = int_0^{\sqrt{2}} \pi ( 4x^2 - x^6 ) \, dx\text{.}\)

4. \(V = \int_0^{2\sqrt{2}} \pi( y^{2/3} - y^2/4 ) \, dy\text{.}\)

5. \(V = \int_0^3 \pi( (y+1)^2 - (y-1)^4 ) \, dy\text{.}\)

1. Observe that horizontal slices with thickness \(\Delta y\) are all cylindrical disks, each with radius given by \(r(y) = y^2\) since they are given by the curve \(y = \sqrt{x}\text{,}\) which can be equivalently expressed as \(x = y^2\text{.}\) The volume of each slice is thus

   \begin{equation*} V_{\text{slice}} = \pi r(y)^2 \Delta y = \pi y^4 \Delta y\text{.} \end{equation*}

   The volume of the solid is thus given by the definite integral

   \begin{equation*} V = \int_0^2 \pi y^4 \Delta \, dy\text{.} \end{equation*}

2. Slicing perpendicular to the axis of rotation, we get slices of thickness \(\Delta y\) that are shaped like washers. The outer radius, \(R(y)\text{,}\) is given by \(R(y) = 4\text{;}\) the innner radius is given by the function \(y = \sqrt{x}\text{,}\) but expressed with \(x\) as a function of \(y\text{,}\) so \(r(y) = y^2\text{.}\) It follows that a representative washer slice has volume

   \begin{equation*} V_{\text{slice}} = \pi (R(y)^2 - r(y)^2) \Delta y = \pi (16 - y^4) \Delta y\text{.} \end{equation*}

   Further, since \(y = \sqrt{x}\) and \(x = 4\) intersect at the point \((4,2)\text{,}\) we integrate from \(y = 0\) to \(y = 2\) and thus find that the total volume is

   \begin{equation*} V = \int_0^2 \pi (16 - y^4) \, dy\text{.} \end{equation*}

3. The curves \(y=x^3\) and \(y=2x\) intersect where \(x^3 = 2x\text{,}\) so where \(x^3-2x = x(x^2 - 2) = 0\text{.}\) The two nonnegative values of \(x\) that make this equation true are \(x=0\) and \(x=\sqrt{2}\text{,}\) and thus the points \((0,0)\) and \((\sqrt{2}, 2\sqrt{2})\) form the boundaries of the region \(S\) that is being rotated about the \(x\)-axis. Since we rotate about the horizontal axis, we slice perpendicular to that axis with slices of thickness \(\Delta x\text{.}\) A representative slice is in the shape of a washer and has volume

   \begin{equation*} V_{\text{slice}} = \pi (R(x)^2 - r(x)^2) \Delta x = \pi ( (2x)^2 - (x^3)^2 ) \Delta x\text{,} \end{equation*}

   and thus the total volume of the solid is

   \begin{equation*} V = int_0^{\sqrt{2}} \pi ( 4x^2 - x^6 ) \, dx\text{.} \end{equation*}

4. To use the same region \(S\) as in part (c), but now revolved about the \(y\)-axis, we shift to horizontal slices of thickness \(\Delta y\text{,}\) and thus need to express the curves as functions of \(y\text{:}\) the rightmost curve is \(x = R(y) = \sqrt[3]{y}\text{,}\) while the leftmost curve is \(x = r(y) = \frac{1}{2}y\text{.}\) A representative slice is a washer whose volume is

   \begin{equation*} V_{\text{slice}} = \pi (R(y)^2 - r(y)^2) \Delta y = \pi( (\sqrt[3]{y})^2 - (\frac{1}{2}y)^2) \Delta y\text{.} \end{equation*}

   It follows that the total volume of the solid is

   \begin{equation*} V = \int_0^{2\sqrt{2}} \pi( y^{2/3} - y^2/4 ) \, dy\text{.} \end{equation*}

5. For this pair of curves, we see they intersect where \(x=(y-1)^2\) and \(x = (y+1)\) are both true, and hence where \((y-1)^2 = y+1\text{.}\) Expanding, moving all terms to the left, and combining like terms, we see that \(y^2 - 2y + 1 = y + 1\text{,}\) so \(y^2 - 3y = 0\text{.}\) It follows that \(y(y-3) = 0\) and therefore \(y=3\) or \(y = 0\text{.}\) Plotting the region and revolving about the \(y\)-axis, we see that horizontal slices of thickness \(\Delta y\) are shaped like washers. The outer radius is given by \(R(y) = y+1\text{,}\) while the inner radius is \(r(y) = (y-1)^2\text{.}\) A representative slice thus has volume

   \begin{equation*} V_{\text{slice}} = \pi (R(y)^2 - r(y)^2) \Delta y = \pi( (y+1)^2 - ((y-1)^2)^2 ) \Delta y\text{.} \end{equation*}

   The volume of the solid of revolution is therefore

   \begin{equation*} V = \int_0^3 \pi( (y+1)^2 - (y-1)^4 ) \, dy\text{.} \end{equation*}

Graphing the region between the two curves in the first quadrant between their points of intersection (\((0,0)\) and \((1,1)\)) and then revolving the region about the line \(y = -1\text{,}\) we see the solid shown in Figure6.27. Each slice of the solid perpendicular to the axis of revolution is a washer, and the radii of each washer are governed by the curves \(y = x^2\) and \(y = x\text{.}\) But we also see that there is one added change: the axis of revolution adds a fixed length to each radius. The inner radius of a typical slice, \(r(x)\text{,}\) is given by \(r(x) = x^2 + 1\text{,}\) while the outer radius is \(R(x) = x+1\text{.}\)

Therefore, the volume of a typical slice is

Finally, we integrate to find the total volume, and

1. Note that the two curves intersect in the first quadrant at \((0,0)\) and \((\sqrt{2}, 2\sqrt{2})\text{.}\)

2. The outer radius is \(R(x) = 4 - x^3\text{.}\)

3. Solve the equations governing the two curves for \(x\) in terms of \(y\text{.}\)

4. The inner radius is \(r(y) = 5 - y^{1/3}\text{.}\)

1. \begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (2x+2)^2 - (x^3 + 2)^2 ) \, dx = \frac{4}{21}(21+8\sqrt{2}) \pi \approx 19.336\text{.} \end{equation*}
2. \begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (4 - x^3)^2 - (4-2x)^2 ) \, dx = \left( 8-\frac{32\sqrt{2}}{21} \right)\pi \approx 18.3626\text{.} \end{equation*}
3. \begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (y^{1/3} + 1)^2 - (\frac{1}{2}y + 1)^2 ) \, dy = \frac{2}{15}(15 + 8\sqrt{2}) \pi \approx 11.022\text{.} \end{equation*}
4. \begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (5 - \frac{1}{2}y)^2 - (5 - y^{1/3})^2 ) \, dy = \frac{2}{15}(75-8\sqrt{2})\pi \approx 26.677\text{.} \end{equation*}

1. The two curves intersect in the first quadrant at \((0,0)\) and \((\sqrt{2}, 2\sqrt{2})\text{.}\) Revolving the region \(S\) about the line \(y = -2\text{,}\) we see that a typical slice is a washer with thickness \(\Delta x\text{,}\) outer radius \(R(x) = 2x + 2\text{,}\) and inner radius \(r(x) = x^3 + 2\text{.}\) It follows by the washer method that

   \begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (2x+2)^2 - (x^3 + 2)^2 ) \, dx = \frac{4}{21}(21+8\sqrt{2}) \pi \approx 19.336\text{.} \end{equation*}

2. In this situation, revolving about \(y = 4\text{,}\) we see that the outer radius of a typical washer is \(R(x) = 4 - x^3\text{,}\) while the inner radius is \(r(x) = 4 - 2x\text{,}\) and thus the volume is

   \begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (4 - x^3)^2 - (4-2x)^2 ) \, dx = \left( 8-\frac{32\sqrt{2}}{21} \right)\pi \approx 18.3626\text{.} \end{equation*}

3. Because we are revolving about \(x = -1\text{,}\) when we slice perpendicular to this we get washers of thickness \(\Delta y\text{,}\) and thus we need to write the two curves as functions of \(y\) in order to determine the radii of the slices as functions of \(y\text{,}\) we note that the line can be expressed as \(x = \frac{1}{2}y\) and the cubic curve can be rewritten in the form \(x = y^{1/3}\text{.}\) The outer radius of a typical slice is thus \(R(y) = y^{1/3} + 1\) and the inner radius is \(r(y) = \frac{1}{2}y + 1\text{.}\) It follows that the volume of the solid is

   \begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (y^{1/3} + 1)^2 - (\frac{1}{2}y + 1)^2 ) \, dy = \frac{2}{15}(15 + 8\sqrt{2}) \pi \approx 11.022\text{.} \end{equation*}

4. Here, the outer radius is \(R(y) = 5 - \frac{1}{2}y\) and the inner radius is \(r(y) = 5 - y^{1/3}\text{,}\) so the total volume of the solid is

   \begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (5 - \frac{1}{2}y)^2 - (5 - y^{1/3})^2 ) \, dy = \frac{2}{15}(75-8\sqrt{2})\pi \approx 26.677\text{.} \end{equation*}

1. Recall that \(L = \int_a^b \sqrt{1+f'(x)^2} \, dx\text{.}\)

2. Remember that \(x^2 + y^2 = 4\) generates a circle centered at \((0,0)\) of radius 4.

3. Apply the arc length formula. Use technology to evaluate the integral.

4. Most expressions involving square roots are difficult to antidifferentiate.

5. Here you can determine the arc length and then experiment to find the upper limit of integration, which will help you determine position.

1. \(L \approx 2.95789\text{.}\)

2. \(L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.}\)

3. \(L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.}\)

4. We will usually have to estimate the value of \(\int_a^b \sqrt{1+f'(x)^2} \, dx\) using computational technology.

5. Approximately \((9.011,f(9.011)) = (9.011, 9.1198)\text{.}\)

1. Using the formula for arc length, we know

   \begin{equation*} L = \int_{-1}^{1} \sqrt{1+(2x)^2} \,dx\text{.} \end{equation*}

   Evaluating the integral, we find \(L \approx 2.95789\text{.}\)

2. The curve is a semi-circle of radius 2, so its length must be \(L = \frac{1}{2} \cdot 2 \pi r = 2\pi\text{.}\) This is confirmed by the arc length formula, since \(f(x) = \sqrt{4-x^2}\text{,}\) so \(f'(x) = \frac{1}{2}(4-x^2)^{-1/2}(-2x)\)

   \begin{equation*} L = \int_{-2}^{2} \sqrt{1 + \left(-x(4-x^2)^{-1/2} \right)^2} \, dx = \int_{-2}^{2} \sqrt{1 + x^2(4-x^2)^{-1}} \, dx\text{.} \end{equation*}

   It follows¹This integral is actually improper because the integrand is undefined at the endpoints, \(x = \pm 2\text{.}\) We learned how to evaluate such integrals in Section5.10. that

   \begin{equation*} L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.} \end{equation*}

3. For \(y = xe^{3x}\text{,}\) we observe that \(y' = 3xe^{3x} + e^{3x} = e^{3x}(3x+1)\text{,}\) and that

   \begin{equation*} (y')^2 = e^{6x}(9x^2 + 6x + 1)\text{.} \end{equation*}

   Thus, the arc length of the curve on \([0,1]\) is

   \begin{equation*} L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.} \end{equation*}

4. Because \(\sqrt{1+f'(x)^2}\) will rarely simplify nicely and rarely have an elementary antiderivative, we will usually have to estimate the value of \(\int_a^b \sqrt{1+f'(x)^2} \, dx\) using computational technology.

5. After 4 seconds traveling at 7 cm/sec, the particle has moved 28 cm. So, we have to find the value of \(b\) for which the arc length \(L\) along \([0,b]\) is 28. Thus, we know that

   \begin{equation*} 28 = L = \int_0^b \sqrt{1 + (0.2x)^2} \, dx\text{.} \end{equation*}

   Rewriting the integral, we see

   \begin{equation*} 28 = \int_0^b \sqrt{1+0.4x^2} \, dx\text{.} \end{equation*}

   Experimenting with different values of \(b\) in a computational engine like Wolfram|Alpha, we find that for \(b \approx 9.011\text{,}\) \(\int_0^{9.011} \sqrt{1+0.4x^2} \, dx \approx 27.9992\text{,}\) and thus the position of the particle when \(t=4\) is approximately \((9.011,f(9.011)) = (9.011, 9.1198)\text{.}\)

1. Use the formula for the arc length for a parametric curve to set up the integral. Recall also the Pythagorean Identity for trig functions: \(\sin^2(t)+\cos^2(t)=1\text{.}\)

2. Use the formula for the arc length for a parametric curve to set up the integral.

1. \(L= \displaystyle \int_0^{2\pi}1 \, dt=2\pi\)

2. \(L= \int_0^{2\pi}\sqrt{(-2\sin(t)^2)+(\cos(t))^2} \, dt \)

1. Since \(\frac{dx}{dt}= -\sin(t)\) and \(\frac{dy}{dt}= \cos(t)\text{,}\) then

   \begin{equation*} L= \int_0^{2\pi}\sqrt{(-\sin(t)^2)+(\cos(t))^2} \, dt = \int_0^{2\pi}1 \, dt=2\pi \end{equation*}

2. Since \(\frac{dx}{dt}= -2\sin(t)\) and \(\frac{dy}{dt}= \cos(t)\text{,}\) then

   \begin{equation*} L= \int_0^{2\pi}\sqrt{(-2\sin(t)^2)+(\cos(t))^2} \, dt = \int_0^{2\pi}\sqrt{4\sin^2(t)+\cos^2(t)} \, dt \end{equation*}

   We do not need to evaluate this integral, and in fact it is an example of an elliptical integral of the second kind, which are a famous class of integrals without elementary antiderivatives.

Set up the integral using formula for arc length for a parametric curve. Then use a computer to evaluate the integral.

\(L \approx 2.266\)

Since \(\frac{dx}{dt}= 3t^2\) and \(\frac{dy}{dt}= 4t^3+1\text{,}\) then

Treat the \(z\)-coordinate the same way the \(x\)- and \(y\)-coordinates are treated.

The length of a 3D line segment is given by \(\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}\text{,}\) so factoring out a factor of \(\Delta t\) and taking the limit as \(\Delta t \rightarrow 0\text{,}\) we get