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Find the area of the surface generated when the given curve is revolved about the given axis: y = x for 0 ≤ x ≤ 1, about the y-axis.

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Final answer:

The surface area generated when the curve y = x is revolved about the y-axis from x = 0 to x = 1 is \(\frac{\pi \sqrt{2}}{2}\) square units.

Step-by-step explanation:

The area of the surface generated by revolving the curve y = x about the y-axis from x = 0 to x = 1 is found using the formula for the surface area of revolution around the y-axis:

\[A = 2\pi \int_{a}^{b} x \sqrt{1 + (\frac{dy}{dx})^2} dx\]

For the curve y = x, the derivative \(\frac{dy}{dx}\) is 1, so the formula simplifies to:

\[A = 2\pi \int_{0}^{1} x \sqrt{2} dx\]

Carrying out the calculation step, we get:

\[A = 2\pi \sqrt{2} \int_{0}^{1} x dx = \pi \sqrt{2} [\frac{1}{2} x^2]_{0}^{1} = \pi \sqrt{2} [\frac{1}{2} (1)^2 - \frac{1}{2} (0)^2]\]

This gives us:

\[A = \pi \sqrt{2} \frac{1}{2} = \frac{\pi \sqrt{2}}{2}\]

The area of the surface generated is \(\frac{\pi \sqrt{2}}{2}\) square units.

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