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Ch. 11 - Parametric Equations and Polar Coordinates
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 11, Problema 11.2.33

Surface Area


Find the areas of the surfaces generated by revolving the curves in Exercises 31-34 about the indicated axes.


x = t + √2, y = (t²/2) + √2t, −√2 ≤ t ≤ √2; y−axis

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1
Identify the parametric equations given: \(x = t + \sqrt{2}\) and \(y = \frac{t^{2}}{2} + \sqrt{2}t\), with the parameter \(t\) ranging from \(-\sqrt{2}\) to \(\sqrt{2}\).
Since the curve is revolved about the \(y\)-axis, use the formula for the surface area of a parametric curve revolved around the \(y\)-axis: \[ S = \int_{a}^{b} 2\pi |x(t)| \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
Compute the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\): - \(\frac{dx}{dt} = 1\) - \(\frac{dy}{dt} = t + \sqrt{2}\)
Substitute \(x(t)\), \(\frac{dx}{dt}\), and \(\frac{dy}{dt}\) into the surface area integral: \[ S = \int_{-\sqrt{2}}^{\sqrt{2}} 2\pi |t + \sqrt{2}| \sqrt{1^2 + (t + \sqrt{2})^2} \, dt \]
Evaluate the integral over the interval \([-\sqrt{2}, \sqrt{2}]\) to find the total surface area. Consider the absolute value in \(|t + \sqrt{2}|\) when setting up the integral, possibly splitting the integral if needed.

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