丁香园AV

Use the method from 厂别肠迟颈辞苍听3.3.1 to find each volume.

1. The base of a tetrahedron (a triangular pyramid) of height \(h\) is an equilateral triangle of side \(s\text{.}\) Its cross-sections perpendicular to an altitude are equilateral triangles. Express its volume \(V\) as an integral, and find a formula for \(V\) in terms of \(h\) and \(s\text{.}\) Verify that your answer is \((1/3)(\hbox{area of base})(\hbox{height})\text{.}\)

   Answer
   \((1/3)(\hbox{area of base})(\hbox{height})\)
   Solution

   We make a diagram below of the base of the tetrahedron:

   

   Therefore, the vertical cross-sections at \(x_i\) are equilateral triangles with side length \(\sqrt{3}x_i\text{.}\) Therefore, the area of this cross-section is

   \begin{equation*} A(x_i) = \frac{\sqrt{3}}{4} \bigl(3 x_i^2\bigr) \end{equation*}

   for \(0 \leq x_i \leq \frac{s}{2}\text{.}\) Note that at \(x_i = s/2\text{,}\) we must have:

   \begin{equation*} h = \frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}s}{4}\right) = \frac{3s}{4}\text{,} \end{equation*}

   which gives the relationship between \(h\) and \(s\text{.}\) We now compute the volume of the solid:

   \begin{equation*} V = 2\int_0^{s/2} A(x) \,dx = 2\int_0^{s/2} \frac{\sqrt{3}}{4} \bigl(3 x^2\bigr)\,dx = \sqrt{3} \frac{s^3}{16}\text{.} \end{equation*}

   We now check that this is equivalent to \(\frac{1}{3}\bigl(\text{ area base } \bigr)h\text{:}\)

   \begin{equation*} \frac{1}{3}\bigl(\text{ area base } \bigr)h = \frac{1}{3} \left(\frac{\sqrt{3}}{4} s^2\right) h= \sqrt{3}\frac{s^3}{16}\text{,} \end{equation*}

   as we found above.

2. The base of a solid is the region between \(f(x)=\cos x\) and \(g(x)=-\cos x\text{,}\) \(-\pi/2\le x\le\pi/2\text{,}\) and its cross-sections perpendicular to the \(x\)-axis are squares. Find the volume of the solid.

   Answer
   \(2\pi\)
   Solution

   We draw a diagram below of the base of the solid:

   

   Therefore, the cross-section at \(x_i\) must be a square with side length \(g(x_i)-f(x_i)\text{.}\) And so the area of this cross-section is

   \begin{equation*} A(x) = \bigl(g(x_i)-f(x_i)\bigr)^2 = 4\cos^2(x_i) \end{equation*}

   for \(0 \leq x_i \leq \frac{\pi}{2}\text{.}\) We now compute the volume of the solid by integrating over these cross-sections:

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

Find the volume of the solid generated by revolving the shaded region about the given axis.

1. About the \(x\)-axis:

   
   Answer
   \(4\pi/3\)
   Solution
   We first write \(y=2-2x\text{.}\) Now integrate:

   \begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[f(x)\right]^2 \,dx \\ \amp=\pi \int_0^1 \left[2-2x\right]^2\,dx \end{split} \end{equation*}

   Let \(u=2-2x\text{,}\) with \(du = -2x\,dx\text{.}\) Then:

   \begin{equation*} \begin{split} V \amp=\pi \int_0^1 \left[2-2x\right]^2\,dx \\ \amp= \pi \int_2^0 \frac{u^2}{2} \,-du\\ \amp= \frac{\pi}{2} \int_0^2 u^2 \,du\\ \amp= \frac{\pi}{6}u^3 \big\vert_0^2 \\ \amp= \frac{4\pi}{3}. \end{split} \end{equation*}

2. About the \(y\)-axis:

   
   Answer
   \(50\pi/3\)
   Solution
   We integrate with respect to \(y\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^2 \pi \left[\frac{5y}{2}\right]^2\,dy \\ \amp= \frac{25\pi}{4}\int_0^2 y^2\,dy \\ \amp= \frac{25\pi}{12} y^3 \big\vert_0^2\\ \amp= \frac{50\pi}{3}. \end{split} \end{equation*}

3. About the \(y\)-axis:

   
   Answer
   \(\pi/2\)
   Solution
   We integrate with respect to \(y\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^1 ]pi \left[\sqrt{y}\right]^2\,dy \\ \amp= \pi \int_0^1 y\,dy \\ \amp= \frac{\pi}{2} y^2 \big\vert_0^1\\ \amp= \frac{\pi}{2}. \end{split} \end{equation*}

4. About the \(x\)-axis:

   
   Answer
   \(\pi^2/32\)
   Solution
   We interate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_{\pi/2}^{\pi/4} \pi\left[\sin x \cos x\right]^2 \,dx \\ \amp= \pi \int_{\pi/2}^{\pi/4} \sin^2 x \cos^2x \,dx \\ \amp= \pi \int_{\pi/2}^{\pi/4} \frac{1-\cos^2(2x)}{4} \,dx \\ \amp= \frac{\pi}{4} \int_{\pi/2}^{\pi/4} \left(1- \frac{1+\cos(4x)}{2}\right)\,dx\\ \amp= -\frac{\pi}{32} \left[\sin(4x)-4x\right]_{\pi/4}^{\pi/2}\\ \amp= \frac{\pi^2}{32}. \end{split} \end{equation*}

Find the volume of the solid generated by revolving the given bounded region about the \(x\)-axis.

1. \(y=x^2, \ y=1, \ x=2, \ y = 0 \)

   Answer
   \(6\pi/5\)
   Solution
   We begin by plotting the area bounded by the given curves:

   

   We split the region into two: one where the top curve is given by \(y=x^2\text{,}\) and the second where the region is bounded by \(y=1\text{.}\) We now use the disk method, and integrate with repect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[x^2\right]^2\,dx + \int_1^2 \pi \left[1\right]^2\,dx \\ \amp= \pi \int_0^1 x^4\,dx + \pi\int_1^2 \,dx \\ \amp= \frac{\pi x^5}{5}\big\vert_0^1 + \pi x \big\vert_1^2\\ \amp= \frac{\pi}{5} + \pi = \frac{6\pi}{5}. \end{split} \end{equation*}

2. \(y=x^3, \ y=0, \ x=1\)

   Answer
   \(\pi/7\)
   Solution
   We first plot the area bounded by the given curves:

   

   We now use the Disk method, and integrate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[x^3\right]^2\,dx \\ \amp= \pi \int_0^1 x^6 \,dx \\ \amp= \left[\frac{\pi x^7}{7}\right]_0^1\\ \amp= \frac{\pi}{7}. \end{split} \end{equation*}

3. \(y=\sqrt{4-x^2}, \ y=0\)

   Answer
   \(32\pi/3\)
   Solution
   We begin by plotting the area bounded by the given curves:

   

   We now use the Disk method, and integrate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_{-2}^2 \pi \left[\sqrt{4-x^2}\right]^2\,dx \\ \amp= \pi \int_{-2}^2 4-x^2\,dx \\ \amp= \pi \left[4x - \frac{x^3}{3}\right]_{-2}^2\\ \amp= \frac{32\pi}{3}. \end{split} \end{equation*}

4. \(y=\sqrt{\sin x}, \ 0\leq x\leq \pi/2\)

   Answer
   \(\pi\)
   Solution
   We begin by plotting the area bounded by the given curves:

   

   We now use the Disk method, and integrate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^{\pi/2} \pi \left[\sqrt{\sin x}\right]^2 \,dx \\ \amp= \pi \int_0^{\pi/2} \sin x \,dx \\ \amp= -\pi \cos x\big\vert_0^{\pi/2}\\ \amp= \pi. \end{split} \end{equation*}

5. \(y=e^{-x}, \ y=0, \ x=2\)

   Answer
   1.5420
   Solution

   We begin by plotting the area bounded by the given curves:

   

   We wish to calculate the volume of the solid generated by rotating this area about the \(x\)-axis. We therefore use the disk method, and integrate with respect to \(x\text{:}\)

   \begin{equation*} V = \int_0^2 \pi (e^{-x})^2 \,dx = \pi \int_0^2 e^{-2x}\,dx = -\frac{\pi}{2}e^{-2x}\bigg\vert_0^2 = -\frac{\pi}{2}\left(e^{-4}-1\right)\text{.} \end{equation*}

6. \(x+y=2, \ y=0, \ x=0\)

   Answer
   \(8\pi/3\)
   Solution
   We begin by plotting the area bounded by the given curves:

   

   We now use the Disk method, and integrate with respect to \(x\text{,}\) and use the substituion \(u=2-x\) with \(du = -dx\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^2 \pi\left[2-x\right]^2\,dx\\ \amp= -\pi \int_2^0 u^2 \,du\\ \amp= \pi \int_0^2 u^2 \,du\\ \amp= \frac{\pi u^3}{3} \bigg\vert_0^2\\ \amp= \frac{8\pi}{3}. \end{split} \end{equation*}

7. \(y=x-x^2, \ y=0, \ x=0\)

   Answer
   \(\pi/30\)
   Solution
   We begin by plotting the area bounded by the curves:

   

   Therefore, we use the Disk method and integrate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[x-x^2\right]^2 \,dx\\ \amp= \pi \int_0^1 x^4-2x^3+x^2 \,dx \\ \amp= \pi \left[\frac{x^5}{5}-\frac{2x^4}{4} + \frac{x^3}{3}\right]_0^1\\ \amp= \pi \left[\frac{1}{5} - \frac{1}{2} + \frac{1}{3} \right]\\ \amp= \frac{\pi}{30}. \end{split} \end{equation*}

8. \(\ds y=\sqrt{\sin x}, \ x=0\)

   Answer
   \(2\pi\)
   Solution
   We notice that \(y=\sqrt(\sin(x)) = 0\) at \(x=\pi\text{.}\) We plot the region below:

   

   We now use the Disk method, and integrate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^{\pi} \pi \left[\sqrt{\sin x}\right]^2 \,dx \\ \amp= \pi \int_0^{\pi} \sin x \,dx \\ \amp= -\pi \cos x\big\vert_0^{\pi}\\ \amp= 2 \pi. \end{split} \end{equation*}

Find the volume of the solid generated by revolving the given bounded region about the \(y\)-axis.

1. \(x=y^2, \ x=0, \ y=-1, \ y=1\)

   Answer
   \(2\pi/5\)
   Solution
   Due to symmetry, the area bounded by the given curves will be twice the green shaded area below:

   

   We therefore use the disk method, and integrate with respect to \(y\text{:}\)

   \begin{equation*} \begin{split} V \amp= 2\int_0^1 \pi \left[y^2\right]^2 \,dy \\ \amp= 2\pi \int_0^1 y^4\,dy \\ \amp= \frac{2\pi y^5}{5} \big\vert_0^1\\ \amp= \frac{2\pi}{5}. \end{split} \end{equation*}

2. \(x=y^{3/2}, \ x=0, \ y=4\)

   Answer
   \(64\pi\)
   Solution
   We begin by plotting the area bounded by the given curves:

   

   We now use the Disk method, and integrate with respect to \(y\text{:}\)

   \begin{equation*} \begin{split} V\amp= \int_0^4 \pi \left[y^{3/2}\right]^2\,dy \\ \amp= \pi \int_0^4 y^3 \,dy \\ \amp= \pi \frac{y^4}{4}\big\vert_0^4 \\ \amp= 64\pi. \end{split} \end{equation*}

3. \(x=\sqrt{\sin(2y)}, \ 0\leq y\leq \pi/2, \ x=0\)

   Answer
   \(\pi\)
   Solution

   The area contained between \(x=0\) and the curve \(x=\sqrt{\sin(2y)}\) for \(0\leq y\leq \frac{\pi}{2}\) is shown below.

   

   Rotating this area around the \(y\)-axis results in a solid with volume

   \begin{equation*} \begin{split} V\amp=\int_0^{\frac{\pi}{2}} \pi \left(\sqrt{\sin(2y)}\right)^2\,dy\\ \amp = \pi\int_0^{\frac{\pi}{2}} \sin(2y)\,dy\\ a\mp = -\frac{\pi}{2}\cos(2y)\bigg\vert_0^{\frac{\pi}{2}}\\ \amp = -\frac{\pi}{2} (-1-1) = \pi.\end{split} \end{equation*}

4. \(x=\sqrt{\cos(2y)},\ 0\leq y\leq \pi/2, \ x=0\)

   Answer
   \(\pi/2\)

Find the volume of the solid generated by revolving the shaded region about the given axis.

1. The \(x\)-axis:

   \begin{equation*} \end{equation*}

   Answer

   \(4\pi \left(\pi-2\right)\)
   Solution
   We notice that the region is bounded on top by the curve \(y=2\text{,}\) and on the bottom by the curve \(y=\sqrt{\cos x}\text{.}\) Therefore, we use the Washer method and integrate with respect to \(x\text{.}\) By symmetry, we have:

   \begin{equation*} \begin{split} V \amp= 2\int_{0}^{\pi/2} \pi \left[2^2 - \left(2\sqrt(\cos x)\right)^2 \right]\,dx\\ \amp= 2\pi \int_{0}^{\pi/2} 4-4\cos x \,dx\\ \amp= 8 \pi \left[x - \sin x\right]_0^{\pi/2}\\ \amp= 4\pi \left(\pi-2\right). \end{split} \end{equation*}

2. The \(y\)-axis:

   \begin{equation*} \end{equation*}

   Answer

   \(\frac{\pi}{4}\left(2\pi-1\right)\)
   Solution
   We notice that the region is bounded on the left by the curve \(x=\sin y\) and on the right by the curve \(x=1\text{.}\) We therefore use the Washer method and integrate with respect to \(y\text{:}\)

   \begin{equation*} \begin{split} V \amp = \int _0^{\pi/2} \pi \left[1 - \sin^2 y\right]\,dy \\ \amp= \pi \int_0^{\pi/2} 1 - \frac{1}{2}\left(1-\cos(2y)\right)\,dy \\ \amp= \frac{\pi}{4}\left(2\pi-1\right). \end{split} \end{equation*}

Find the volume of the solid generated by revolving the given bounded region about the \(x\)-axis.

1. \(y=x, \ y=2, \ x=0\)

   Answer
   \(16\pi/3\)
   Solution
   We first plot the area bounded by the given curves:

   

   We notice that the area is bounded above by the curve \(y=2\) for \(x\in[0,2]\text{,}\) and bounded below by the curve \(y=x\) for \(x\in[0,2]\text{.}\) Therefore, we use the Washer method and integrate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^2 \pi \left[2^2-x^2\right]\,dx\\ \amp= \pi \int_0^2 \left[4-x^2\right]\,dx\\ \amp= \pi\left[4x-\frac{x^3}{3}\right]_0^2\\ \amp=\frac{16\pi}{3}. \end{split} \end{equation*}

2. \(y=3\sqrt{x}, \ y=3, \ x=0\)

   Answer
   \(9\pi/2\)
   Solution
   We first plot the area bounded by the given curves:

   

   We notice that the region is bounded above by the curve \(y=3\) for \(x\in[0,1]\) and bounded below by the curve \(y=3\sqrt{x}\) for \(x\in[0,1]\text{.}\) Therefore, we use the Washer method and integrate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[3^2-\bigl(3\sqrt{x}\bigr)^2\right]\,dx\\ \amp= \pi \int_0^1 \left[9-9x\right]\,dx\\ \amp= \pi\left[9x-\frac{9x^2}{2}\right]_0^1\\ \amp=\frac{9\pi}{2}. \end{split} \end{equation*}

3. \(y=x^2+1,\ y+x=3\)

   Answer
   \(117\pi/5\)
   Solution

   The points of intersection of the curves \(y=x^2+1\) and \(y+x=3\) are calculated to be

   \begin{equation*} \begin{gathered} x^2+1=3-x \\ x^2+x-2 = 0 \\ (x-1)(x+2) = 0 \\ \implies x=1,-2. \end{gathered} \end{equation*}

   Let \(f(x)=x^2+1\) and \(g(x)=3-x\text{.}\) We now plot the area contained between the two curves:

   

   To find the volume of the solid which is generated when this area is rotated around the \(x\)-axis, we use the washer method, and integrate with respect to \(x\text{.}\) We find

   \begin{equation*} V = \int_{-2}^1 \pi\left[(3-x)^2 - (x^2+1)^2\right]\,dx = \pi \left[-\frac{x^5}{5} - \frac{x^3}{3} - 3x^2 + 8x\right]_{-2}^1 = \frac{117\pi}{5}\text{.} \end{equation*}

4. \(y=9-x^2, \ y=3-x\)

   Answer
   \(125/3 (6\pi-1)\)
   Solution
   We first compute the intersection point(s) of the two curves:

   \begin{equation*} \begin{gathered} x^2+1=3-x \\ x^2-x-6 = 0 \\ (x-3)(x+2) = 0 \\ \implies x=3,-2. \end{gathered} \end{equation*}

   We now plot the area contained between the two curves:

   

   The region is bounded above by the curve \(y=9-x^2\) for \(x\in[-2,3]\text{,}\) and bounded below by the curve \(y=3-x\) for \(x\in[-2,3]\text{.}\) We therefore use the Washer method and integrate with respect to \(x\text{:}\)

   \begin{equation*} \begin{split} V \amp= \int_{-2}^3 \pi \left[(9-x^2)^2 - (3-x)^2\right)\,dx \\ \amp= \pi \int_{-2}^3 \left[x^4-19x^2+6x+72\right]\,dx\\ \amp = \pi \left[\frac{x^5}{5}-19\frac{x^3}{3} + 3x^2 + 72x\right]_{-2}^3\\ \amp= \frac{125}{3}\bigl(6\pi-1\bigr) \end{split} \end{equation*}

The equation \(\ds x^2/9+y^2/4=1\) describes an ellipse. Find the volume of the solid obtained by rotating the ellipse around the \(x\)-axis and also around the \(y\)-axis. These solids are called ellipsoids; one is vaguely rugby-ball shaped, one is sort of flying-saucer shaped, or perhaps squished-beach-ball-shaped.

For the first solid, we consider the following region:

Notice that for this region, we have \(y = 2\sqrt{1-\frac{x^2}{9}}\text{.}\) Therefore, we use the Disk method and integrate with respect to \(x\text{:}\)

\begin{equation*} \begin{split} V \amp= \int_{-3}^3 \pi \left[2\sqrt{1-\frac{x^2}{9}}\right]^2\,dx \\ \amp= 4\pi \int_{-3}^3 \left(1-\frac{x^2}{9}\right)\,dx\\ \amp= 4\pi \left[x - \frac{x^3}{9(3)}\right]_{-3}^3\\ \amp= 16 \pi. \end{split} \end{equation*}

Next, we consider the following region:

Now we write \(x = 3\sqrt{1-\frac{y^2}{4}}\) and integrate with respect to \(y\text{:}\)

\begin{equation*} \begin{split} V \amp= \int_{-2}^2 \pi \left[3\sqrt{1-\frac{y^2}{4}}\right]^2\,dy \\ \amp= 9\pi \int_{-2}^2 \left(1-\frac{y^2}{4}\right)\,dx\\ \amp= 9\pi \left[x - \frac{y^3}{4(3)}\right]_{-2}^2\\ \amp= 24 \pi. \end{split} \end{equation*}

Use integration to compute the volume of a sphere of radius \(r\text{.}\) You should of course get the well-known formula \(\ds 4\pi r^3/3\text{.}\)

Consider the region the curve \(y^2+x^2=r^2\) such that \(y \geq 0\text{:}\)

Notice that the shaded region is bounded above by the curve \(y = \sqrt{r^2-x^2}\) for \(x \in[-r,r]\) and below by the \(x\)-axis. Therefore, to find the volume of the sphere of radius \(r\text{,}\) we use the Disk method and integrate with respect to \(x\text{:}\)

\begin{equation*} \begin{split} V \amp= \int_{-r}^r \pi \left[\sqrt{r^2-x^2}\right]^2\,dx\\ \amp= \pi \int_{-r}^r \left(r^2-x^2\right)\,dx\\ \amp= \pi \left[r^2 x - \frac{x^3}{3}\right]_{-r}^r \\ \amp= \pi \left[\left(r^3-\frac{r^3}{3}\right)-\left(-r^3+\frac{r^3}{3}\right)\right]\\ \amp= \pi \left(2r^3-\frac{2r^3}{3}\right)\\ \amp= \frac{4\pi r^3}{3}, \end{split} \end{equation*}

which is exactly the formula for the volume of a sphere of radius \(r\text{.}\)

A hemispheric bowl of radius \(r\) contains water to a depth \(h\text{.}\) Find the volume of water in the bowl.

The bowl can be described as the solid obtained by rotating the following region about the \鈥�(y\)-axis:

(Note: this is the part of the ellipse \(\frac{y^2}{h^2} + \frac{x^2}{r^2} = 1\) for which \(y \geq 0\) and \(0\leq x\leq r\)). Therefore, to find the volume of the solid, we use the Disk method and integrate with respect to \(y\text{:}\)

\begin{equation*} \begin{split} V\amp= \int_{0}^h \pi \left[r\sqrt{1-\frac{y^2}{h^2}}\right]^2\, dy\\ \amp= \pi r^2 \int_0^h \left(1-\frac{y^2}{h^2}\right)\,dy\\ \amp= \frac{2}{3}\pi h r^2 \end{split} \end{equation*}