Upward Flux Over a Nonrectangular Projection
For a graph surface, the height above each point in the plane is specified by a function. The surface’s projection onto the \(xy\)-plane may be a triangle, a region between curves, or a shape that needs to be divided into pieces. The flux calculation uses that projected region as its domain of integration.
The central task in this tutorial is therefore twofold: form the integrand for the upward orientation, then choose bounds that describe exactly the projection of the surface. Incorrect bounds can produce a perfectly evaluated integral for the wrong portion of the surface.
The Upward Graph Flux Integrand
Write a vector field as \(\mathbf{F}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle\). A parametrization of the graph is \(\mathbf{r}(x,y)=\langle x,y,g(x,y)\rangle\). Its tangent vectors are \(\mathbf{r}_x=\langle 1,0,g_x\rangle\) and \(\mathbf{r}_y=\langle 0,1,g_y\rangle\). The cross product \(\mathbf{r}_x\times\mathbf{r}_y=\langle -g_x,-g_y,1\rangle\) has positive \(z\)-component, so it points upward.
This formula explains why the projected region matters. The oriented area vector accounts for the graph’s slope, while \(dA\) measures area in the \(xy\)-plane. Once the integrand has been formed, the remaining integral is over \(D\), using bounds suited to that region.
Write the planar region \(D\) as bounds in \(x\) and \(y\), or divide it into simpler pieces.
Substitute \(z=g(x,y)\) into the field, then take its dot product with \(\langle -g_x,-g_y,1\rangle\).
Use the chosen bounds for \(D\), and add the integrals if the region was divided into nonoverlapping pieces.
Worked Example: Flux Above a Triangular Region
Worked Example: Flux Above a Triangular Region
Find the upward flux of \(\mathbf{F}(x,y,z)=\langle x,y,2z\rangle\) across \(z=x+y\) above the triangle with vertices \((0,0)\), \((1,0)\), and \((0,1)\).
The triangle is described by \(0\le x\le1\) and \(0\le y\le1-x\). The graph function is \(g(x,y)=x+y\), so \(g_x=1\) and \(g_y=1\). On the graph, the field is \(\langle x,y,2(x+y)\rangle\). Its dot product with the upward area vector is \[ \langle x,y,2(x+y)\rangle\cdot\langle -1,-1,1\rangle =-x-y+2(x+y)=x+y. \] Thus the flux is \[ \int_0^1\int_0^{1-x}(x+y)\,dy\,dx. \]
First integrate with respect to \(y\): \[ \int_0^{1-x}(x+y)\,dy =\left[xy+\frac{y^2}{2}\right]_0^{1-x} =x(1-x)+\frac{(1-x)^2}{2}. \] Expanding gives \(x-x^2+\frac12-x+\frac{x^2}{2}=\frac12-\frac{x^2}{2}\). Therefore, \[ \int_0^1\left(\frac12-\frac{x^2}{2}\right)dx =\left[\frac{x}{2}-\frac{x^3}{6}\right]_0^1 =\frac12-\frac16 =\frac13. \]
As a check, the integral of \(x+y\) over the triangle is the sum of the integrals of \(x\) and \(y\). They are \[ \int_0^1 x(1-x)\,dx=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac16 \] and \[ \int_0^1\int_0^{1-x}y\,dy\,dx =\frac12\int_0^1(1-x)^2\,dx =\frac12\left[x-x^2+\frac{x^3}{3}\right]_0^1 =\frac16. \] Their sum is \(\frac16+\frac16=\frac13\), agreeing with the flux calculation.
Choosing Bounds Between Curves
A common way to describe an irregular region is to give the range of one coordinate first, then specify the lower and upper curves for the other coordinate. For example, if \(a\le x\le b\), and the region lies between \(y=u(x)\) and \(y=v(x)\), its bounds are \(a\le x\le b\), \(u(x)\le y\le v(x)\). The inner limits must follow the actual bottom and top of the region for every \(x\) in the stated interval.
Worked Example: Flux Above a Region Bounded by Curves
Find the upward flux of \(\mathbf{F}(x,y,z)=\langle 0,0,z\rangle\) across \(z=x^2+y\) above the region \(D\) bounded by \(y=x^2\), \(y=1\), and \(x=0\).
The curves \(y=x^2\) and \(y=1\) meet when \(x^2=1\). Together with the boundary \(x=0\), the specified region has \(0\le x\le1\), with \(x^2\le y\le1\). Here \(g=x^2+y\), so \(g_x=2x\) and \(g_y=1\). Since the first two components of the field are zero, the upward flux integrand is just its third component on the graph: \(R(x,y,g(x,y))=g(x,y)=x^2+y\). Hence \[ \text{Flux}=\int_0^1\int_{x^2}^{1}(x^2+y)\,dy\,dx. \]
Integrating in \(y\) gives \[ \left[x^2y+\frac{y^2}{2}\right]_{x^2}^{1} =x^2+\frac12-x^4-\frac{x^4}{2} =x^2+\frac12-\frac32x^4. \] Then \[ \int_0^1\left(x^2+\frac12-\frac32x^4\right)dx =\left[\frac{x^3}{3}+\frac{x}{2}-\frac{3x^5}{10}\right]_0^1 =\frac13+\frac12-\frac{3}{10} =\frac{8}{15}. \]
For a check, separate the two terms in the integrand. The \(x^2\) contribution is \[ \int_0^1x^2(1-x^2)\,dx =\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1 =\frac{2}{15}. \] The \(y\) contribution is \[ \int_0^1\int_{x^2}^{1}y\,dy\,dx =\frac12\int_0^1(1-x^4)\,dx =\frac12\left(1-\frac15\right) =\frac25. \] The total is \(\frac{2}{15}+\frac25=\frac{2}{15}+\frac{6}{15}=\frac{8}{15}\), as found above.
When the Region Must Be Split
Some projections are easiest to describe as a union of simpler regions. If the pieces do not overlap except along boundaries, integrate over each piece and add the results. A shared boundary has planar area zero, so it does not contribute an extra area term. The split should make the bounds accurate and manageable; it should not include the same interior area twice.
Worked Example: Splitting an L-Shaped Projection
Find the upward flux of \(\mathbf{F}(x,y,z)=\langle 0,0,z\rangle\) across \(z=x+y\) above the L-shaped region \[ D=([0,2]\times[0,1])\cup([0,1]\times[0,2]). \]
The two rectangles overlap on \([0,1]\times[0,1]\), so they cannot both be integrated as written and then added: that would count the overlap twice. Instead, divide the L into nonoverlapping pieces: \(D_1=[0,1]\times[0,2]\) and \(D_2=[1,2]\times[0,1]\). The upward integrand is \(R(x,y,g(x,y))=x+y\). Thus \[ \text{Flux} =\int_0^1\int_0^2(x+y)\,dy\,dx +\int_1^2\int_0^1(x+y)\,dy\,dx. \]
For the first piece, \[ \int_0^2(x+y)\,dy=2x+2, \qquad \int_0^1(2x+2)\,dx=[x^2+2x]_0^1=3. \] For the second piece, \[ \int_0^1(x+y)\,dy=x+\frac12, \qquad \int_1^2\left(x+\frac12\right)dx =\left[\frac{x^2}{2}+\frac{x}{2}\right]_1^2 =3-1=2. \] Therefore the flux is \(3+2=5\).
As a check, use horizontal slices instead. For \(0\le y\le1\), \(x\) runs from \(0\) to \(2\); for \(1\le y\le2\), \(x\) runs from \(0\) to \(1\). The same integral is \[ \int_0^1\int_0^2(x+y)\,dx\,dy +\int_1^2\int_0^1(x+y)\,dx\,dy. \] The first inner integral is \(2+2y\), whose integral from \(0\) to \(1\) is \(3\). The second is \(\frac12+y\), whose integral from \(1\) to \(2\) is \(\frac32+\frac12=2\). The total is again \(5\).
A Common Bounds Error
The bounds must describe the projection, not merely a convenient rectangle containing it. For the triangular region in the first example, the incorrect calculation \[ \int_0^1\int_0^1(x+y)\,dy\,dx \] uses the entire unit square. It gives \[ \int_0^1\left(x+\frac12\right)dx =\frac12+\frac12=1, \] which is not the flux through the triangular patch. The square includes points with \(x+y>1\), which are outside the projection.
The correction is to use the sloping boundary of the triangle: \(0\le x\le1\), \(0\le y\le1-x\). With those limits the integral is \(\frac13\), as computed earlier. Before evaluating, test the proposed bounds against the region’s boundaries and check that each interior point is included once and only once.
Check Your Understanding
For each question, identify the projection bounds before evaluating any flux integral.
- What upward area vector is used for the graph \(z=g(x,y)\), and why does it represent the upward orientation?
- Describe the bounds for the triangle \(0\le x\), \(0\le y\), \(x+y\le2\) using \(x\) as the outer variable.
- For the region \(0\le x\le2\), \(x\le y\le2\), and graph \(z=g(x,y)\), write the iterated integral for upward flux of \(\mathbf{F}=\langle0,0,1\rangle\).
- Why must overlapping pieces be handled carefully when splitting a planar projection?
- What error is made by using an enclosing rectangle as the integration region for a triangular projection?