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Mathematics · Ch 9 — Applications of Integration

Volume of a Solid Obtained by Revolving Area About an Axis

9.9

Volume of a Solid Obtained by Revolving Area About an Axis

Solids of revolution. When a plane region is given one complete rotation (360∘=2π360^\circ=2\pi radians) about a fixed axis lying in its own plane, it sweeps out a solid of revolution. For example, revolving the semicircular region enclosed by x2+y2=a2x^2+y^2=a^2 above the xx-axis about the xx-axis generates a sphere of radius aa; revolving the rectangular region bounded by y=0, y=a, x=0, x=hy=0,\,y=a,\,x=0,\,x=h about the xx-axis generates a right-circular cylinder of radius aa and height hh.

This section restricts to revolution about the xx-axis or the yy-axis. For revolution about the xx-axis, the revolved plane region lies above the xx-axis (y≥0y\ge0); for revolution about the yy-axis, it lies to the right of the yy-axis (x≥0x\ge0).

Derivation (disc method), about the xx-axis. Let y=f(x)y=f(x), xx-axis, x=a, x=bx=a,\,x=b (b>ab>a) bound a region in the first quadrant, with every vertical line between x=ax=a and x=bx=b meeting the curve exactly once. Divide [a,b][a,b] into nn segments x0=a<x1<⋯<xn=bx_0=a<x_1<\cdots<x_n=b, Δx=b−an\Delta x=\frac{b-a}n. On each subinterval, the thin rectangle of height yi=f(xi)y_i=f(x_i) and width Δx\Delta x, revolved about the xx-axis, sweeps out an elementary cylindrical disc of radius yiy_i and height Δx\Delta x, hence volume πyi2Δx\pi y_i^2\Delta x (using "volume of a cylinder =πr2h=\pi r^2h"). Summing all the discs, ∑πyi2 Δx\sum \pi y_i^2\,\Delta x, and letting n→∞, Δx→0n\to\infty,\ \Delta x\to0, this tends to the volume of the whole solid:

V=π∫aby2 dx(revolution about the x-axis).\boxed{V=\pi\int_a^b y^2\,dx} \qquad \text{(revolution about the }x\text{-axis).}

By the identical argument with xx and yy interchanged, for a curve x=f(y)x=f(y), yy-axis, and y=c, y=dy=c,\,y=d revolved about the yy-axis,

V=π∫cdx2 dy(revolution about the y-axis).\boxed{V=\pi\int_c^d x^2\,dy} \qquad \text{(revolution about the }y\text{-axis).}

Standard solids re-derived by these formulas (Examples 9.62-9.69), all worth having on hand as checks:

  • Sphere of radius aa: revolve y=a2−x2y=\sqrt{a^2-x^2}, −a≤x≤a-a\le x\le a, about the xx-axis: V=π∫−aa(a2−x2) dx=2π∫0a(a2−x2) dx=43πa3V=\pi\int_{-a}^a(a^2-x^2)\,dx=2\pi\int_0^a(a^2-x^2)\,dx=\dfrac43\pi a^3.
  • Right circular cone, base radius rr, height hh: revolve the triangular region under y=rhxy=\dfrac rh x, 0≤x≤h0\le x\le h, about the xx-axis: V=π∫0h(rhx)2dx=13πr2hV=\pi\int_0^h\left(\dfrac rhx\right)^2dx=\dfrac13\pi r^2h.
  • Spherical cap of height hh cut from a sphere of radius rr: revolve y=r2−x2y=\sqrt{r^2-x^2}, r−h≤x≤rr-h\le x\le r, about the xx-axis: V=πh2 ⁣(r−h3)V=\pi h^2\!\left(r-\dfrac h3\right); in terms of the cap's own base radius ρ\rho (where ρ2+(r−h)2=r2\rho^2+(r-h)^2=r^2), V=πh6(3ρ2+h2)V=\dfrac{\pi h}{6}\left(3\rho^2+h^2\right).
  • Ellipsoid, from the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 (a>ba>b) revolved about the major axis (xx-axis): V=43πab2V=\dfrac43\pi ab^2; revolved about the minor axis (yy-axis) instead: V=43πa2bV=\dfrac43\pi a^2b. …
Figure 9.34The upper semicircular region inside x^2+y^2=a^2 above the x-axis, bounded by x=-a and x=a, which when revolved one full turn about the x-axis generates a sphere.
Fig. 9.34 — The upper semicircular region inside x^2+y^2=a^2 above the x-axis, bounded by x=-a and x=a, which when revolved one full turn about the x-axis generates a sphere.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The upper semicircular region inside x^2+y^2=a^2 above the x-axis, bounded by x=-a and x=a, which when revolved one full turn about the x-axis gene …

Figure 9.35The rectangular plane region bounded by y=0, y=a, x=0 and x=h which, when revolved one full turn about the x-axis, generates a right-circular cylinder of radius a and height h.
Fig. 9.35 — The rectangular plane region bounded by y=0, y=a, x=0 and x=h which, when revolved one full turn about the x-axis, generates a right-circular cylinder of radius a and height h.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The rectangular plane region bounded by y=0, y=a, x=0 and x=h which, when revolved one full turn about the x-axis, generates a right-circular cylinder of radiu …

Figure 9.36The plane region under y=f(x) between x=a and x=b, with one elementary vertical strip of width delta x, revolved about the x-axis to form a thin cylindrical disc.
Fig. 9.36 — The plane region under y=f(x) between x=a and x=b, with one elementary vertical strip of width delta x, revolved about the x-axis to form a thin cylindrical disc.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The plane region under y=f(x) between x=a and x=b, with one elementary vertical strip of width delta x, revolved about the x-axis to form a thin cy …

Figure 9.37The plane region bounded by x=f(y), the y-axis and the lines y=c and y=d, with one elementary horizontal strip of width delta y, revolved about the y-axis.
Fig. 9.37 — The plane region bounded by x=f(y), the y-axis and the lines y=c and y=d, with one elementary horizontal strip of width delta y, revolved about the y-axis.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The plane region bounded by x=f(y), the y-axis and the lines y=c and y=d, with one elementary horizontal strip of width delta y, revolved abou …

Figure 9.38The semicircular region bounded by y=sqrt(a^2-x^2) and the x-axis between x=-a and x=a revolved about the x-axis to generate a sphere of radius a, drawn with meridian and equatorial cross-sections.
Fig. 9.38 — The semicircular region bounded by y=sqrt(a^2-x^2) and the x-axis between x=-a and x=a revolved about the x-axis to generate a sphere of radius a, drawn with meridian and equatorial cross-sections.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The semicircular region bounded by y=sqrt(a^2-x^2) and the x-axis between x=-a and x=a revolved about the x-axis to generate a sphere of radius a, drawn with meridian and equato …

Figure 9.39The triangular region in the first quadrant bounded by y=(r/h)x, the x-axis and x=h, revolved about the x-axis to generate a right-circular cone of base radius r and height h.
Fig. 9.39 — The triangular region in the first quadrant bounded by y=(r/h)x, the x-axis and x=h, revolved about the x-axis to generate a right-circular cone of base radius r and height h.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The triangular region in the first quadrant bounded by y=(r/h)x, the x-axis and x=h, revolved about the x-axis to generate a right-circular cone of base radius …

Figure 9.40The region in the first quadrant bounded by the circle x^2+y^2=r^2, the x-axis and the lines x=r-h and x=r, revolved about the x-axis to generate a spherical cap of height h.
Fig. 9.40 — The region in the first quadrant bounded by the circle x^2+y^2=r^2, the x-axis and the lines x=r-h and x=r, revolved about the x-axis to generate a spherical cap of height h.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The region in the first quadrant bounded by the circle x^2+y^2=r^2, the x-axis and the lines x=r-h and x=r, revolved about the x-axis to generate a spherical …

Figure 9.41The region bounded by the parabola y=x^2, the x-axis and the ordinates x=0 and x=1, revolved about the x-axis to generate a paraboloidal solid.
Fig. 9.41 — The region bounded by the parabola y=x^2, the x-axis and the ordinates x=0 and x=1, revolved about the x-axis to generate a paraboloidal solid.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The region bounded by the parabola y=x^2, the x-axis and the ordinates x=0 and x=1, revolved about the x-axis to generate a parabol …

Figure 9.42The region bounded by the ellipse x^2/a^2 + y^2/b^2 = 1 (a>b) revolved about the major axis (x-axis) to generate a prolate ellipsoid; vertices (a,0), (-a,0) and (0,b).
Fig. 9.42 — The region bounded by the ellipse x^2/a^2 + y^2/b^2 = 1 (a>b) revolved about the major axis (x-axis) to generate a prolate ellipsoid; vertices (a,0), (-a,0) and (0,b).

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The region bounded by the ellipse x^2/a^2 + y^2/b^2 = 1 (a>b) revolved about the major axis (x-axis) to generate a prolate ellipsoid; vertices (a,0), (- …

Figure 9.43The region bounded by the parabola x=y^2+1, the y-axis and the lines y=1 and y=-1, revolved about the y-axis to generate a solid of revolution.
Fig. 9.43 — The region bounded by the parabola x=y^2+1, the y-axis and the lines y=1 and y=-1, revolved about the y-axis to generate a solid of revolution.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The region bounded by the parabola x=y^2+1, the y-axis and the lines y=1 and y=-1, revolved about the y-axis to generate a solid of …

Figure 9.44The region bounded by the hyperbola portion y=(3/4)sqrt(x^2-16) (that is x^2/16 - y^2/9 = 1), the y-axis and the lines y=1 and y=6, revolved about the y-axis.
Fig. 9.44 — The region bounded by the hyperbola portion y=(3/4)sqrt(x^2-16) (that is x^2/16 - y^2/9 = 1), the y-axis and the lines y=1 and y=6, revolved about the y-axis.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The region bounded by the hyperbola portion y=(3/4)sqrt(x^2-16) (that is x^2/16 - y^2/9 = 1), the y-axis and the lines y=1 and y=6, revolved abo …

Figure 9.45The region bounded by the curve y=log x, the lines y=0, x=0 and y=2, revolved about the y-axis.
Fig. 9.45 — The region bounded by the curve y=log x, the lines y=0, x=0 and y=2, revolved about the y-axis.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The region bounded by the curve y=log x, the lines y=0, x=0 and y=2, revolved about the y-a …

Figure 9.46A container in the shape of a right-circular conical frustum with top radius 2 m, bottom radius 1 m and height 2 m (Exercise 9.9, Q5).
Fig. 9.46 — A container in the shape of a right-circular conical frustum with top radius 2 m, bottom radius 1 m and height 2 m (Exercise 9.9, Q5).

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. A container in the shape of a right-circular conical frustum with top radius 2 m, bottom radius 1 m and height 2 m (Exercise …