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AI HL · AHL 3

Further Integration

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AHL 3.10

Harder volumes and areas of revolution

AHL 3.10

Harder volumes and areas of revolution

Target: I can find an unknown limit when the volume of revolution is given, by writing the volume as a function of the limit and solving. (+5 more)

By the end of this section

Key points

Which formula, which variable

  • About the x-axis: \(V=\pi\int y^2\,dx\). The limits are x-values.
  • About the y-axis: \(V=\pi\int x^2\,dy\). Write x in terms of y, and use y-values as limits. A limit given as an x-value must be turned into a y-value.
  • Between two curves: \(V=\pi\int\left(\text{outer}^2-\text{inner}^2\right)dx\). Square each curve, then subtract. Never square the difference.
  • A region between a curve and the y-axis rotates about the y-axis with no hole. A region under a curve, away from the y-axis, leaves a hole: subtract a cylinder or the inner solid.
  • A glass has an outer and an inner surface. The capacity is the volume inside the inner surface, from the base of the inside up to the top. The glass itself is the volume inside the outer surface minus the capacity.

Unknown limits and exact values

  • To find an unknown limit \(a\): integrate with the limit left as \(a\), set the result equal to the given volume, then use Solve for \(a\) on the calculator.
  • Cancel \(\pi\) from both sides before you solve. Check the answer by putting it back into the volume.
  • Exact answers keep \(\pi\), \(e\) and roots: write \(\frac{48\pi}{5}\), not 30.2. If the question says "exact", use the calculator's exact mode.
  • Intersection points of two curves are the limits of the integral. Write the points as coordinates when asked.
  • Rotation about a line that is not an axis is not in this course. A translated curve is: translate first, then rotate about the axis.

Showing results by integration

  • Cone: a line through the origin, \(y=\frac rh x\), rotated about the x-axis from 0 to \(h\), gives \(V=\pi\int_0^h\frac{r^2}{h^2}x^2dx=\frac13\pi r^2h\).
  • Sphere: the circle \(x^2+y^2=r^2\) gives \(y^2=r^2-x^2\). Rotating the upper half gives \(V=\pi\int_{-r}^{r}(r^2-x^2)dx=\frac43\pi r^3\). Change the limits for a cap.
  • To show a result, write every line. Your last line must be exactly the result you were given.
  • Area in terms of y: \(\int x\,dy\). A rectangle's area = the area under the curve + the area beside it. This is how you show \(\int\ln x\,dx\) and ratios of areas.

On the calculator

A definite integral for a volume

  • Enter the integrand with the square included, such as \((x^3+8)\) or \(\left(x\right)^2\), and the limits.
  • Multiply by \(\pi\) after the integral if the integrand has no \(\pi\).
  • Give 3 significant figures, or use exact mode when the question asks for an exact value.

Screenshot and video to come

Solving for an unknown limit

  • Write the volume equation with the limit as a, for example \(\pi(1-1/a)=4\pi/5\).
  • Use Solve for a, with a positive starting value. Check that it makes sense in the question.

Screenshot and video to come

Intersections of two curves

  • Graph both curves and find the points where they meet.
  • Use the x-values of the intersections as the limits of the integral.

Screenshot and video to come

Intercepts, area and volume of one curve

  • Graph the curve. Find the x-intercept with the zero tool.
  • Find the area with a definite integral between the intercepts, and the volume with the squared integral times \(\pi\).

Screenshot and video to come

Explore: a solid of revolution

Drag the two limits to choose the region. Turn it about the axis and watch each strip sweep out a disc. Then change the number of discs and compare their total with the exact volume.



Limits
Sum of the discs
Exact volume
The sum of the pieces is

The solid is drawn as seen from a little to one side, so each circle looks like an ellipse. Each disc takes its radius from the middle of its strip.

Turn the region to see the discs.

  1. Set θ = 0, then turn slowly to 360°. What does one strip become? Why is its volume π y² δx and not π y δx?
  2. Set n = 1, then 5, then 20. What happens to the sum of the discs? Predict, then show the exact volume.
  3. Choose y = 2/x about the x-axis with a = 1 and show the exact volume. Drag b until it is 2π, about 6.28. Check your b by solving π∫ (2/x)² dx = 2π from 1 to b.
  4. Choose Washers and turn the region 360°. Each strip now leaves a hole, so its volume is π(outer² − inner²) δx. Make the hole bigger. Does the volume go up or down? Predict, then show the exact volume.
  5. Turn y = x² + 1 about the y-axis. Why are the limits now y-values, and why does the working use x² = y − 1?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Finding an unknown limit

(a) The part of the curve \(y=\frac1x\) between \(x=1\) and \(x=a\) (where \(a>1\)) is rotated 360° about the x-axis. The volume is \(\frac{4\pi}{5}\). Find a. (b) The part of \(y=x^2\) between \(x=0\) and \(x=b\) is rotated about the y-axis. The volume is \(6\pi\). Find b.

Working
  1. Volume as a function of a

    \(V=\pi\int_1^a\frac1{x^2}\,dx=\pi\left[-\frac1x\right]_1^a=\pi\left(1-\frac1a\right)\)

    Square y first. The unknown limit stays in the answer.

  2. Set equal to the given volume

    \(\pi\left(1-\frac1a\right)=\frac{4\pi}{5}\)

    Cancel π: the equation is 1 − 1/a = 4/5.

  3. Solve for a

    Solve for a: a = 5

    The calculator solves it. Check: π(1 − 1/5) = 4π/5.

  4. Part (b): about the y-axis

    \(x^2=y,\quad V=\pi\int_0^{b^2}y\,dy=\frac{\pi b^4}{2}=6\pi\)

    About the y-axis the limits are y-values: 0 to b². So b⁴ = 12.

  5. Solve for b

    Solve for b: b = 1.86

    b is positive, so take the positive root.

Video walkthrough coming
Rotating about the y-axis, with a cylinder

The region is bounded by the curve \(y=\sqrt{x-2}\), the line \(x=11\) and the x-axis. It is rotated 360° about the y-axis. Find the volume to the nearest integer.

24681012−11234xyy = √(x − 2)
Working
  1. Write x in terms of y

    \(x=y^2+2\)

    About the y-axis the integral is in y. At x = 11, y = 3.

  2. The cylinder

    \(\pi\times11^2\times3=363 \pi\)

    Radius 11 (the line x = 11), height 3.

  3. The part to remove

    \(\pi\int_0^3(y^2+2)^2\,dy=\frac{483 \pi}{5}\)

    This is the solid between the y-axis and the curve. The region we want is outside it.

  4. Subtract

    \(V=\frac{1332\pi}{5}=836.920\)

    To the nearest integer, V = 837.

Video walkthrough coming
Area, then volume about each axis

The curve \(y=\sqrt{x^3+8}\) meets the axes at \(A(-2,0)\) and \(B(0,2\sqrt2)\). The region R is bounded by the curve and the two axes. (a) Show that \(x^2=\sqrt[3]{(y^2-8)^2}\). (b) Find the area of R. (c) Find the volume when R is rotated about (i) the x-axis, (ii) the y-axis.

−3−2−112−11234xyy = √(x³ + 8)AB
Working
  1. (a) Rearrange for x

    \(y^2=x^3+8\;\Rightarrow\;x^3=y^2-8\;\Rightarrow\;x^2=(y^2-8)^{2/3}=\sqrt[3]{(y^2-8)^2}\)

    Square the curve, move the 8, then raise to the power 2/3. This is the form you need for the y-axis.

  2. (b) Area

    From the calculator: \(\int_{-2}^{0}\sqrt{x^3+8}\,dx=4.76\)

    Use the x-form for the area. It cannot be done by hand here.

  3. (c)(i) About the x-axis

    \(V=\pi\int_{-2}^0(x^3+8)\,dx=12 \pi\) = 37.7

    Squaring a square root removes it, so this integral is exact.

  4. (c)(ii) About the y-axis

    From the calculator: \(V=\pi\int_{0}^{2\sqrt2}\sqrt[3]{(y^2-8)^2}\,dy=26.3\)

    Use the y-form from part (a). The y-limits are 0 and 2√2.

Video walkthrough coming
Volume between two curves

Find the volume formed when the region between \(y=x^3\) and \(y=\sqrt x\) is rotated 360° about the x-axis.

0.510.51xyy = √xy = x³O
Working
  1. Intersections

    \(x^3=\sqrt x\;\Rightarrow\;x^6=x\;\Rightarrow\;x=0\text{ or }1\)

    The limits come from where the curves meet. Check with the graph: (0, 0) and (1, 1).

  2. Outer and inner

    \(\text{outer: }y=\sqrt x,\quad\text{inner: }y=x^3\)

    On 0 < x < 1 the root is above the cube.

  3. The integral

    \(V=\pi\int_0^1\left(x-x^6\right)dx\)

    Outer squared minus inner squared: (√x)² − (x³)² = x − x⁶.

  4. Evaluate

    From the calculator: V = \(\frac{5 \pi}{14}\) = 1.12

    Give 3 significant figures.

Video walkthrough coming
The volume of a cone from a line

The line through \((5,0)\) and \((0,12)\) is rotated 360° about the y-axis, from \(y=0\) to \(y=12\). It forms a cone. (a) Find the equation of the line in the form \(ax+by=c\). (b) Find the volume of the cone.

−112345672468101214xy(5, 0)(0, 12)
Working
  1. (a) The line

    \(\frac xy\text{ form: }\frac x5+\frac y{12}=1\;\Rightarrow\;12x+5y=60\)

    The intercepts are 5 and 12, so x/5 + y/12 = 1. Multiply by 60.

  2. x in terms of y

    \(x=\frac{60-5y}{12}\)

    About the y-axis we need x² as a function of y.

  3. (b) Integrate

    \(V=\pi\int_0^{12}\left(\frac{60-5y}{12}\right)^2dy=100\pi\)

    From the calculator: 100π. Check with the cone formula: ⅓π × 5² × 12 = 100π.

  4. Give the number

    V = 314

    100π = 314 to 3 significant figures.

Video walkthrough coming
Checking the volume of a sphere

A circle has equation \(x^2+y^2=36\). (a) Show that the upper half has equation \(y=\sqrt{36-x^2}\). (b) The upper half is rotated 360° about the x-axis. Show that the volume of the sphere formed is \(288\pi\), and check this against the formula \(\frac43\pi r^3\).

−8−6−4−22468−6−4−2246xy
Working
  1. (a) Upper half

    \(y^2=36-x^2\;\Rightarrow\;y=\sqrt{36-x^2}\quad(y\ge0)\)

    Take the positive root: the upper half.

  2. (b) The integral

    \(V=\pi\int_{-6}^{6}\left(36-x^2\right)dx=\pi\left[36x-\frac{x^3}3\right]_{-6}^{6}\)

    The semicircle runs from x = −6 to x = 6.

  3. Evaluate

    \(=\pi\left(144-(-144)\right)=288\pi\)

    216 − 72 = 144 at x = 6; the value at −6 is −144.

  4. Check

    \(\frac43\pi\times6^3=288\pi\)

    The two agree. In decimals V = 905.

Video walkthrough coming
An area that gives an integral

The region R is bounded by the curve \(y=\ln x\), the coordinate axes and the line \(y=\ln 6\). (a) Find the area of R. (b) Hence show that \(\int_1^6\ln x\,dx=6\ln6-5\).

−112345678−2−1123xyy = ln xln 6
Working
  1. (a) Integrate in y

    \(x=e^y,\quad\text{Area}=\int_0^{\ln6}e^y\,dy=\left[e^y\right]_0^{\ln6}=6-1=5\)

    R lies between the y-axis and the curve, so integrate x with respect to y.

  2. (b) The rectangle

    \(6\times\ln6=6\ln6\)

    The rectangle from the origin to (6, ln 6) is R plus the area under the curve from 1 to 6.

  3. Subtract R

    \(\int_1^6\ln x\,dx=6\ln6-5\)

    Numerically 6 ln 6 − 5 = 5.75.

Video walkthrough coming
A glass: capacity and material

The cross-section of a glass is shaded in the diagram. The outer surface is \(y=0.2x^2\) and the inner surface is \(y=1.5+0.15x^2\), for \(x\ge0\). All lengths are in cm. The two surfaces meet at the rim. (a) Find the height of the glass. (b) The glass is rotated 360° about the y-axis. Find the volume of liquid the glass can hold. (c) Find the volume of glass used.

1234561234567x (cm)y (cm)
Working
  1. Where they meet

    \(0.2x^2=1.5+0.15x^2\Rightarrow 0.05x^2=1.5\Rightarrow x^2=30\), so \(y=0.2\times30=6\)

    The height of the glass is 6 cm.

  2. x² in terms of y

    Outer: \(x^2=5y\). Inner: \(x^2=\frac{y-1.5}{0.15}\)

    Rotating about the y-axis, so write x² in terms of y for both surfaces.

  3. Capacity

    \(V=\pi\int_{1.5}^{6}\frac{y-1.5}{0.15}\,dy=212\) cm³

    Liquid fills inside the inner surface, from the inside base at y = 1.5 up to the rim.

  4. The glass itself

    Outer solid: \(\pi\int_0^{6}5y\,dy=90\pi=282.7\). Capacity: \(67.5\pi=212.1\). Glass: \(22.5\pi=70.7\) cm³

    Volume inside the outer surface, minus the capacity.

Video walkthrough coming