Find the area of the surface. The part of the paraboloid $z = 1 - x^2 - y^2$ that lies above the plane $z = -4$.

Neetesh Kumar | November 27, 2024

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This is the solution to Math 1D
Assignment: 15.5 Question Number 4
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Step-by-step solution:

To compute the surface area of the given portion of the paraboloid, we follow these steps:

Step 1: Understand the surface and boundary

The equation of the paraboloid is $z = 1 - x^2 - y^2$. The plane $z = -4$ intersects the paraboloid, defining the lower boundary of the region of interest. To find this boundary, set $z = -4$:

$1 - x^2 - y^2 = -4$

Simplify:

$x^2 + y^2 = 5$

Thus, the region of interest lies above the plane $z = -4$ within the disk defined by $x^2 + y^2 \leq 5$ in the $xy$-plane.

Step 2: Surface area formula

The formula for the surface area $S$ of a parametric surface $z = f(x, y)$ is:

$S = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dA$

Here, $f(x, y) = 1 - x^2 - y^2$. Compute the partial derivatives:

$\frac{\partial f}{\partial x} = -2x, \quad \frac{\partial f}{\partial y} = -2y$

Substitute these into the formula:

$S = \iint_R \sqrt{1 + (-2x)^2 + (-2y)^2} \, dA$

Simplify:

$S = \iint_R \sqrt{1 + 4x^2 + 4y^2} \, dA$

Step 3: Convert to polar coordinates

Since the region $R$ is a disk defined by $x^2 + y^2 \leq 5$, it is natural to use polar coordinates. In polar form:

$x = r\cos\theta, \quad y = r\sin\theta, \quad dA = r \, dr \, d\theta$

Also, $x^2 + y^2 = r^2$. Substituting these into the surface area formula:

$S = \int_0^{2\pi} \int_0^{\sqrt{5}} \sqrt{1 + 4r^2} \, r \, dr \, d\theta$

Step 4: Simplify the integral

Separate the integral:

$S = \int_0^{2\pi} d\theta \int_0^{\sqrt{5}} r \sqrt{1 + 4r^2} \, dr$

The $\theta$-integral evaluates to:

$\int_0^{2\pi} d\theta = 2\pi$

Thus:

$S = 2\pi \int_0^{\sqrt{5}} r \sqrt{1 + 4r^2} \, dr$

Step 5: Solve the radial integral

Let $u = 1 + 4r^2$, so $du = 8r \, dr$. When $r = 0$, $u = 1$. When $r = \sqrt{5}$, $u = 1 + 4(5) = 21$.

Rewriting the integral in terms of $u$:

$\int_0^{\sqrt{5}} r \sqrt{1 + 4r^2} \, dr = \frac{1}{8} \int_1^{21} \sqrt{u} \, du$

The integral of $\sqrt{u}$ is:

$\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$

Evaluate:

$\frac{1}{8} \int_1^{21} \sqrt{u} \, du = \frac{1}{8} \left[\frac{2}{3} u^{3/2} \right]_1^{21}$

Substitute the limits:

$= \frac{1}{8} \left[\frac{2}{3} (21)^{3/2} - \frac{2}{3} (1)^{3/2} \right]$

Simplify:

$= \frac{1}{8} \cdot \frac{2}{3} \left(21^{3/2} - 1 \right)$

$= \frac{1}{12} \left(21^{3/2} - 1 \right)$

Step 6: Final surface area

Multiply by $2\pi$ to get the total surface area:

$S = 2\pi \cdot \frac{1}{12} \left(21^{3/2} - 1 \right)$

Simplify further:

$S = \frac{\pi}{6} \left(21^{3/2} - 1 \right)$

Thus, the surface area of the portion of the paraboloid is:

$S = \frac{\pi}{6} \left(21^{3/2} - 1 \right)$

Final Answer:

The area of the surface is: $S = \boxed{\frac{\pi}{6} \left(21^{3/2} - 1 \right)}$

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