Question

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. $\iint (\mathrm{curl}\mathbf{V})\cdot \mathbf{n}d\sigma$ over the part of the surface $z=9-x^2-9y^2$ above the $(x,y)$ plane if $\mathbf{V}=2xy\mathbf{i}+(x^2-2x)\mathbf{j}-x^2{z}^2\mathbf{k}$

Short answer

The result is $-6\cdot \pi$.

Step-by-step solution

Understand the Integral

The integral to compute is $\iint (\mathrm{curl}\mathbf{V})\cdot \mathbf{n}d\sigma$ over the surface given by $z=9-x^2-9y^2$. The vector field is $\mathbf{V}=2xy\mathbf{i}+(x^2-2x)\mathbf{j}-x^2{z}^2\mathbf{k}$.

Choose the Appropriate Theorem

Since the integral involves the curl of a vector field over a surface, applying Stokes' Theorem is appropriate. Stokes' Theorem relates the surface integral of curl of $\mathbf{V}$ over a surface to a line integral around the boundary curve of the surface: ${\iint }_S(\mathrm{curl}\mathbf{V})\cdot \mathbf{n}d\sigma ={\oint }_{\partial S}\mathbf{V}\cdot d\mathbf{r}$.

Describe the Surface and its Boundary

The surface is given by $z=9-x^2-9y^2$ above the $xy$-plane. This is a part of a paraboloid opening downwards. The boundary of this surface is a circle on the $xy$-plane where $z=0$, i.e., $9-x^2-9y^2=0$, which simplifies to $$x^2+9y^2=9.$$

Parameterize the Boundary Curve

Parameterize the boundary curve $C$. Set $x=3\cos t$ and $y=\sin t$ for $0\le t\le 2\pi$. In vector form, $\mathbf{r}(t)=3\cos t\mathbf{i}+\sin t\mathbf{j}$.

Compute $\mathbf{V}\cdot d\mathbf{r}$

First, find $d\mathbf{r}$. Differentiating $\mathbf{r}(t)$ gives $d\mathbf{r}=(-3\sin t)dt\mathbf{i}+(\cos t)dt\mathbf{j}$. Then, substitute $x=3\cos t$ and $y=\sin t$ into $\mathbf{V}$ giving $$\mathbf{V}=2(3\cos t)(\sin t)\mathbf{i}+((3\cos t{)}^2-2(3\cos t))\mathbf{j}.$$ Simplify to $$\mathbf{V}=6\cos t\sin t\mathbf{i}+(9{\cos }^{2}t-6\cos t)\mathbf{j}.$$

Calculate the Line Integral

The line integral becomes $${\oint }_C\mathbf{V}\cdot d\mathbf{r}={\int }_0^{2\pi }(6\cos t\sin t\cdot (-3\sin t)+(9{\cos }^{2}t-6\cos t)\cos t)dt.$$ Simplify to $${\int }_0^{2\pi }(-18\cos t{\sin }^{2}t+9{\cos }^{3}t-6{\cos }^{2}t)dt.$$ Each term can be processed using trigonometric identities or known integrals.

Evaluate Each Integral Term

Separate the integral into three parts: $${\int }_0^{2\pi }-18\cos t{\sin }^{2}tdt,{\int }_0^{2\pi }9{\cos }^{3}tdt,{\int }_0^{2\pi }-6{\cos }^{2}tdt.$$ Using periodicity and symmetry properties of sine and cosine, note most terms vanish over $0$ to $2\pi$. One integral simplifies to $${\int }_0^{2\pi }-6{\cos }^{2}tdt=-6\cdot \pi .$$

Key concepts

Surface Integral

A surface integral extends the concept of an integral to functions defined on a surface. It can be thought of as the summation of quantities across a surface in three-dimensional space. In the context of vector fields, a surface integral can be computed by integrating the dot product of the vector field and the normal to the surface over the entire surface area. This is mathematically expressed as ${\iint }_S\mathbf{F}\cdot \mathbf{n}d\sigma$, where $\mathbf{F}$ is the vector field, $\mathbf{n}$ is the normal vector, and $d\sigma$ represents a surface element. Understanding surface integrals is essential for applying Stokes' Theorem, which relates the integral of the curl of a vector field over a surface to a line integral around the boundary of the surface.

Curl of a Vector Field

The curl of a vector field is a vector operation that describes the infinitesimal rotation at any given point in the field. For a vector field $\mathbf{V}=P\mathbf{i}+Q\mathbf{j}+R\mathbf{k}$, the curl is given by $\mathrm{curl}\mathbf{V}=\text{abla}\times \mathbf{V}$. This is calculated using the determinant of a matrix involving the unit vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ and the partial derivatives of the functions $P,Q,R$. It results in a new vector field that indicates the rotation per unit area at each point: $$abla\times \mathbf{V}=(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})\mathbf{i}-(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z})\mathbf{j}+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathbf{k}$$. Calculating the curl helps determine the behavior of the vector field, particularly in fluid dynamics and electromagnetism.

Line Integral

A line integral computes the integral of a function along a curve. When dealing with vector fields, this specifically involves integrating the dot product of the vector field and a differential element along a path. Mathematically, for a vector field $\mathbf{F}$ and a curve parameterized by $\mathbf{r}(t)$, the line integral is given by: $${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\int }_a^b\mathbf{F}(\mathbf{r}(t))\cdot \frac{d\mathbf{r}(t)}{dt}dt$$. This integral effectively measures the cumulative effect of the vector field along the path. In the context of Stokes' Theorem, the line integral around the boundary curve of a surface relates to the surface integral of the curl of the vector field over that surface. Understanding and computing line integrals are crucial in applying this theorem.

Parameterization of Boundary Curve

Parameterizing the boundary curve involves expressing the curve as a vector function of a single parameter, typically $t$, which varies over a certain interval. For a curve $C$ in the plane or in space, parameterization simplifies calculations in integrals. When dealing with the boundary of surfaces, parameterizing the curve allows for easier conversion of line integrals. For example, consider a circle defined by $x^2+9y^2=9$. By setting $x=3\cos t$ and $y=\sin t$ with $0\le t\le 2\pi$, we get a smooth path representation where each point on the boundary can be easily described as $\mathbf{r}(t)=3\cos t\mathbf{i}+\sin t\mathbf{j}$. Correct parameterization converts complex integrals into more manageable forms and is critical in verifying the application of Stokes' Theorem.