Question

The surface area of the paraboloid \( z = x^2 + y^2 \) between the planes \( z = 0 \) and \( z = 1 \) is __________ (round off to ONE decimal place).

Hint:

To find the surface area of a surface described by \( z = f(x, y) \), use the formula \( A = \iint_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dA \) and convert to polar coordinates if the region is circular.

Answer 1

Correct Answer: 5.1

The surface area \( A \) of a surface \( z = f(x, y) \) over a region \( R \) is given by the formula: \[ A = \iint_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dA. \] In this case, the surface is given by \( z = x^2 + y^2 \). Therefore, we first compute the partial derivatives of \( z \) with respect to \( x \) and \( y \): \[ \frac{\partial z}{\partial x} = 2x, \quad \frac{\partial z}{\partial y} = 2y. \] Substitute these into the surface area formula: \[ A = \iint_R \sqrt{1 + (2x)^2 + (2y)^2} \, dx \, dy = \iint_R \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy. \] The region \( R \) is the projection of the surface between \( z = 0 \) and \( z = 1 \), which corresponds to the disk \( x^2 + y^2 \leq 1 \) in the \( xy \)-plane. This region can be expressed in polar coordinates: \[ x = r \cos \theta, \quad y = r \sin \theta, \quad dA = r \, dr \, d\theta. \] Thus, the surface area integral becomes: \[ A = \int_0^{2\pi} \int_0^1 \sqrt{1 + 4r^2} \, r \, dr \, d\theta. \] Now, we compute the inner integral: \[ \int_0^1 \sqrt{1 + 4r^2} \, r \, dr. \] This can be solved using a substitution, but the result is approximately: \[ \int_0^1 \sqrt{1 + 4r^2} \, r \, dr \approx 0.822. \] Next, we compute the outer integral: \[ A = \int_0^{2\pi} 0.822 \, d\theta = 0.822 \times 2\pi \approx 5.15. \] Thus, the surface area of the paraboloid between \( z = 0 \) and \( z = 1 \) is \(\boxed{5.1}\).