Vector Calculus

Under Pressure: Surface Integrals

Combining vectors together with calculus concepts creates new kinds of derivatives and integrals, many of which we'll explore through their use in the physical world.

To get a sense of what's to come, we'll use fluid mechanics problems to introduce the surface integral, an important extension of the double integral of multivariable calculus. We'll also investigate the Divergence Theorem in this setting in the sequel.

No prior physics experience is required, however.

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

Today's our first day apprenticing at Tanks For All The Fish, a business of building aquariums. No job is too big or too small, and no tank shape is too weird or too difficult.

Our first task is to build a small cubic fish tank \( l \) units on a side.

According to our boss, the renowned Mr. D. Adams, the total force exerted on the container walls is an important quantity to keep in mind when choosing building materials. The force on a small patch of a container wall of area \( \triangle A \) is given by the pressure \( p_{_\text{fluid}} \) times the area, and is directed perpendicularly outward along a unit normal \(\hat{n}.\)

Find the total force vector on the bottom of the tank, assuming the pressure is constant there and using the coordinate system displayed here:

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

To find the total force vector on the bottom of the cubic aquarium tank, we assumed constant fluid pressure.

We got away with this only because the fluid is in hydrostatic equilibrium (i.e. it's not sloshing or moving around) and every point on the bottom of the tank is at the same depth below the water.

Mr. Adams has made a visualization demonstrating a very important rule in the aquarium building business:

The pressure on any small patch of the container wall is proportional to the depth below the surface, and the fluid force is directed perpendicularly outward.

The visualization above shows one such small patch of the container wall and the total force (blue arrow) exerted on it by the fluid it's holding back (invisible, but on the side opposite the blue arrow). Assume that \(z\) points vertically up and that the patch's center is located at \( (a,b,c).\) The angles \(\theta\) and \(\phi\) control the patch's orientation. Notice how the length of the vector (strength of the force) changes with \( c,\) the depth below the surface.

We'll have to take this into account if we want to compute instead the total fluid force on one of the side walls of the cubic aquarium.

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

Mr. Adams tasks us with computing the total force vector on the side wall located at \( x = l.\)

The pressure below the surface of a full tank is \[ p(z) = p_{_\text{fluid}}\left [ 1-\frac{z}{l} \right], \] which expresses mathematically the idea that the pressure is proportional to depth.

Let's set up a double integral that determines the total force vector on the \( x = l \) side of the tank.

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

Now let's compute the total force vector on the \( x = l \) side of the cubic aquarium tank: \[ \vec{F}_{\text{tot}, x = l} = \left( \iint\limits_{[0,l]\times[0,l]} p_{_\text{fluid}}\left [ 1-\frac{z}{l} \right]dy\, dz \right) \hat{i}.\] Is the magnitude of this force greater than, equal to, or less than that for the bottom of the tank? \[ \vec{F}_{\text{bottom}} = - p_{_\text{fluid}} l^2 \hat{k} \]

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

Mr. Adams's motto is there's no aquarium shape too weird or too difficult. To initiate us into advanced aquarium design, we're next tasked with constructing a hemispherical tank, a slightly more challenging shape than a cube.

We can approximate the total fluid force just as before, but we'll need to first find the outward-pointing unit normal vector at the point \[ \left (x,y,l-\sqrt{l^2-x^2-y^2} \right ) \] on the hemispherical wall. Here, \( l \) is the radius.

What is this vector?


It's displayed below; sliders \( a \) and \( b \) control the position of point \( \left (a,b, l-\sqrt{l^2-a^2-b^2} \right ) \) (all red), and the green arrow represents \( \hat{n},\) the outward-pointing unit normal vector there.

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

It's still true that the pressure is proportional to depth, and the force is directed perpendicularly outwards even though the wall is now hemispherical.

If we break the hemisphere up into very small patches, the total force is approximately \[ \vec{F}_{\text{tot}} \approx \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} p(z_{ij}^{*}) \triangle A_{ij} \hat{n}_{ij} = p_{_\text{fluid}}\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} \left [ 1- \frac{z_{ij}^{*}}{l} \right] \triangle A_{ij} \hat{n}_{ij}. \] We know \(\hat{n}_{ij}, \) so we just need to deal with \( \triangle A_{ij},\) the area of patch \( (i,j). \)

The patch (red) sits above a very small rectangle (blue) in the \( xy\)-plane of dimensions \( \triangle x_{i} \) by \( \triangle y_j \) and centered on \( a=x_{i}^{*},\ b =y_{j}^{*}.\) Our Riemann sum for the total fluid force will become a double integral in \( x \) and \(y,\) but we need to figure out how \( \triangle A_{ij} \) relates to \( \triangle x_{i} \triangle y_{j}\) first.

The picture exaggerates to make things easier to visualize, but the hemispherical patch is approximately a parallelogram with four vertices \[ \left( x_{i}^{*} \pm \frac{\triangle x_{i}}{2} , \ y_{j}^{*} \pm \frac{\triangle y_{j}}{2}, f \bigg( x_{i}^{*} \pm \frac{\triangle x_{i}}{2} , \ y_{j}^{*} \pm \frac{\triangle y_{j}}{2}\bigg) \right), \] and \( f = l - \sqrt{l^2-x^2-y^2}.\) Since \( \triangle x_{i} \) and \( \triangle y_{j}\) are very, very small, we can approximate \[ f \bigg( x_{i}^{*} \pm \frac{\triangle x_{i}}{2} , \ y_{j}^{*} \pm \frac{\triangle y_{j}}{2}\bigg) \approx f \left( x_{i}^{*} , \ y_{j}^{*} \right) \pm \frac{\triangle x_{i}}{2}f_{x}(x_{i}^{*},y_{j}^{*})\pm \frac{\triangle y_{j}}{2}f_{y}(x_{i}^{*},y_{j}^{*}).\] From this we gather that two vectors defining the parallelogram patch are \[ \begin{align} \vec{s}_{1} & = \left \langle \triangle x_{i}, 0 , \triangle x_{i} \ f_{x}(x_{i}^{*},y_{j}^{*}) \right \rangle, \ \vec{s}_{2} = \left \langle 0 , \triangle y_{j}, \triangle y_{j} \ f_{y}(x_{i}^{*},y_{j}^{*}) \right \rangle, \end{align} \] and so the area of this parallelogram (i.e. the patch area) is \[ \triangle A_{ij} \approx \| \vec{s}_{1} \times \vec{s}_{2} \| = \sqrt{1+\big[f_{x}(x_{i}^{*},y_{j}^{*})\big]^2+\big[f_{y}(x_{i}^{*},y_{j}^{*})\big]^2} \triangle x_{i} \triangle y_{j}. \]


Learn more about first-order approximations and their extensions on our Taylor Series wiki.

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

As we take the size of the patches infinitesimally small, the total force Riemann sum approaches the double integral \[ \begin{align} \vec{F}_{\text{tot}} & = p_{_\text{fluid}} \iint\limits_{x^2+y^2 \leq l^2} \left [ 1-\frac{f(x,y)}{l} \right] \hat{n}\, d A, \end{align} \] where we abbreviate \( d A = \sqrt{ 1+[f_{x}]^2+[f_{y}]^2}\, dx\, dy \) and \[ f(x,y) = l - \sqrt{l^2-x^2-y^2},\ \ \hat{n} =\left \langle \frac{x}{l} , \frac{y}{l} , \frac{f(x,y)}{l}-1 \right \rangle. \] Let's compute the \(\hat{i}\)-component of the total fluid force.


Hint: Think about the symmetry of the problem to reduce the amount of integration needed.

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

We found that the only nonzero component of the total fluid force is in the \( \hat{k}\)-direction. Symmetry provided a quick route to the answer, which was confirmed through a direct integration.

As a consequence of our previous calculations, we have \[ \vec{F}_{\text{tot}} = -p_{_\text{fluid}} \left( \iint\limits_{x^2+y^2\leq l^2} \sqrt{1-\left( \frac{x^2+y^2}{l^2} \right)}\ dx\, dy \right) \hat{k}.\] Use polar coordinates \( x = r \cos(\theta), \ y = r \sin(\theta) \) and the identity \[ \int\limits_{u=0}^{u=1} u \sqrt{1-u^2}\, du = \frac{1}{3} \] to find the total fluid force on the hemispherical tank.

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   

Under Pressure: Surface Integrals

We're well on our way to becoming a master aquarium builder; all it takes is a little vector calculus!

Finding the total fluid force on the hemispherical tank \( S\) amounted to computing \[ \vec{F}_{\text{tot}} = \iint_{S} p \ \hat{n}\, dA,\] a new kind of integral over a surface, or surface integral. As our course develops, we'll see a great need for integrating both functions and vector fields over nice surfaces like the hemispherical tank.

Because of the complexity and importance of surface integrals, we'll have a whole chapter devoted to them. That same chapter also builds from scratch the Divergence Theorem, one of the most important results of our course.

In the next unit, we'll use what we learned apprenticing for Mr. Adams at Tanks For All The Fish to begin to understand the depth of this remarkable theorem.

Under Pressure: Surface Integrals

Intro to Vector Calculus

                   
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