Question

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint_{\text {surface } \sigma} \operatorname{curl}\left(x^{2} \mathbf{i}+z^{2} \mathbf{j}-y^{2} \mathbf{k}\right) \cdot \mathbf{n} d \sigma,\) where \(\sigma\) is the part of the surface \(z=4-x^{2}-y^{2}\) above the \((x, y)\) plane.

Short answer

The value of the integral is \(\frac{16}{3}\).

Step-by-step solution

Understand the Given Integral

The integral to evaluate is \ \(\iint_{\text{surface } \sigma} \operatorname{curl}(x^{2} \mathbf{i} + z^{2} \mathbf{j} - y^{2} \mathbf{k}) \cdot \mathbf{n} d \sigma\). The surface \(\sigma\) is part of the paraboloid \(z = 4 - x^{2} - y^{2}\) above the \((x, y)\) plane.

Apply Stokes' Theorem

Stokes' Theorem states that \ \(\iint_{\text{surface } \sigma} \operatorname{curl} \mathbf{F} \cdot \mathbf{n} d \sigma = \oint_{C} \mathbf{F} \cdot d \mathbf{r}\), where \(C\) is the boundary of \(\sigma\). The boundary of \(\sigma\) is the circle \(x^{2} + y^{2} = 4\) in the \(z = 0\) (\(x, y\)) plane.

Parameterize the Boundary Curve

Parameterize the circle \(C\) using \ \(\mathbf{r}(t) = 2 \cos t \mathbf{i} + 2 \sin t \mathbf{j}\), where \(t\) ranges from 0 to \(2\pi\). The differential \(d\mathbf{r} = -2 \sin t \mathbf{i} + 2 \cos t \mathbf{j} dt\).

Evaluate \(\mathbf{F} \cdot d\mathbf{r}\)

Evaluate the line integral by substituting the parameterization into \(\mathbf{F}\). Given \(\mathbf{F} = x^{2} \mathbf{i} + z^{2} \mathbf{j} - y^{2} \mathbf{k}\), and noting that on the boundary curve \(z = 0\), \ \(\mathbf{F}(t) = 4 \cos^{2} t \mathbf{i} - 4 \sin^{2} t \mathbf{k}\). Therefore, only the \(\mathbf{i}\)-component contributes: \ \(\mathbf{F} \cdot d\mathbf{r} = 4 \cos^{2} t (-2 \sin t dt) = -8 \cos^{2} t \sin t dt\).

Integrate Over the Boundary Curve

Integrate \ \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} = \int_{0}^{2\pi} -8 \cos^{2} t \sin t dt\). To simplify this integral, use the substitution \(u = \cos t\), hence \(du = -\sin t dt\). The integral becomes: \ \(\int_{1}^{-1} 8 u^{2} du = 8 \int_{-1}^{1} u^{2} du = 16 \int_{0}^{1} u^{2} du\).

Compute the Final Value

Evaluating the integral, \ \(16 \int_{0}^{1} u^{2} du = 16 \left[ \frac{u^{3}}{3} \right]_{0}^{1} = 16 \cdot \frac{1}{3} = \frac{16}{3}\). Thus, the value of the integral is \(\frac{16}{3}\).

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics focused on vector fields and operations on them, such as differentiation and integration. It's divided into various parts, one of which deals with differential operators like the gradient, divergence, and curl. These operators help in understanding the behavior of vector fields. For instance, in our exercise, we deal with the curl of a vector field. The curl operation measures the rotation or the 'twisting' of the field. This is crucial for applying Stokes' theorem, as the theorem relates the curl of a vector field over a surface to the line integral of the field over the boundary of the surface.

Surface Integrals

Surface integrals extend the concept of integrals to two-dimensional surfaces. They are useful for calculating quantities that spread over a surface, such as flux. To compute a surface integral, we need a vector field and a surface. In Stokes' theorem, the surface integral of the curl of a vector field over a surface can be converted into a line integral over the surface's boundary. For our exercise, the surface is the part of the paraboloid above the \((x, y) \) plane, and the vector field is \(x^{2} \mathbf{i} + z^{2} \mathbf{j} - y^{2} \mathbf{k}\).

Paraboloid

A paraboloid is a quadratic surface that can be used to model various physical shapes and phenomena. In our problem, the surface \(z = 4 - x^{2} - y^{2}\) describes an upward-facing paraboloid. This surface is significant because it informs the boundary of the region over which the surface integral is computed. The boundary is the circle \(x^{2} + y^{2} = 4\) lying in the \(z = 0\) plane. By parameterizing this boundary, we were able to convert the surface integral into a more manageable line integral, ultimately solving the problem using Stokes' theorem.