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How to Calculate Shells Calculus

One of the most immediate applications of integral calculus is the determination of volume for non-standard shapes. These shapes are often represented by a function which is then rotated about an axis of the Cartesian plane. The area under the curve is then extrapolated using one of several methods. This first method often taught to students is the cylindrical shell method. This method creates infinitely thin cylindrical shells to represent the area of a curve between two points of integration.

Instructions

    • 1

      Determine the appropriate limits of integration. These limits are values that bound the real number line under the curve. The limits of integration are either explicitly stated in the problem or, if performed outside of a classroom setting, are equal to the distance on the real number line taken up by the absolute length of the line being integrated. For example, in finding the volume of a parabola, set the limits of integration between the points where the parabola crosses the x-axis. It is between these two points, or limits, that the integration takes place and the cylindrical shells are created.

    • 2

      Set up a standard integral of the form ∫ x * f(x) dx. Where ∫ is the standard integration symbol and "dx" is the standard mathematical notation for "with respect to x." "With respect to x" is used due to the limits of integration being along the x-axis.

    • 3

      Multiply the integral by the constant, pi. Because this value is a constant, it may be multiplied against the entire integral itself. From the above step, calculate 2π ∫ x * f(x) dx.

    • 4

      Evaluate the integral at the limits of integration. Use the fundamental theorem of calculus to create an anti-derivative for the function (see Tips and Resources).

    • 5

      Evaluate the anti-derivative at each limit of integration.

    • 6

      Subtract the value at the second limit of integration, the limit whose value is largest, from the initial limit of integration. For example, if the limits of integration were 0 and 15, you would evaluate the anti-derivative at 15, then subtract the value of the anti-derivative, calculated at 0.

    • 7

      Multiply the resulting value by 2π. This value of 2π may be left outside the evaluated integral as the distributive property will make it a common factor. The resulting value is the volume of the function between the defined limits.

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