Section 3.3 Week 3 Activity
Subsection 3.3.1 Jacobians
Activity 3.3.1.
Calculate the Jacobian of this transformation.
First, I need to calculate the partial derivatives.
Then I put the partial derivative into the Jacobian matrix and take the determinant to calculate the Jacobian.
Activity 3.3.2.
Calculate the Jacobian of this transformation.
First, I need to calculate the partial derivatives.
Then I put the partial derivative into the Jacobian matrix and take the determinant to calculate the Jacobian.
Activity 3.3.3.
Calculate the Jacobian of this transformation.
First, I need to calculate the partial derivatives.
Then I put the partial derivative into the Jacobian matrix and take the determinant to calculate the Jacobian.
Activity 3.3.4.
Calculate the Jacobian of this transformation.
First, I need to calculate the partial derivatives.
Then I put the partial derivative into the Jacobian matrix and take the determinant to calculate the Jacobian.
Subsection 3.3.2 Change of Variables
Activity 3.3.5.
Calculate the integral using a change of variables. (There is a transformation which will simplify both the region of integration and the integrand.)
The region of integration \(D\) is the parallelogram with vertices \((0,0)\text{,}\) \((4,-1)\text{,}\) \((3,2)\text{,}\) \((7,1)\text{,}\) as show in Figure 3.3.1.
It would simplify the integrand if \(u = 2x-3y\) and \(v = 4y+x\text{.}\) If I solve the system to invert this transformation, I get \(x = 3u+4v\) and \(y = 2u-v\text{.}\) Furthermore, if I look at the line that bound the parallelogram, I get the lines \(y = \frac{2}{3}x\text{,}\) \(y = \frac{3}{2} x - \frac{11}{3}\text{,}\) \(y = \frac{-1}{4} x\) and \(y = \frac{-1}{4} x + \frac{11}{4}\text{.}\) If I substitute for \(u\) and \(v\) in these bounds, I get the bounds (respective) \(u=0\text{,}\) \(u=11\text{,}\) \(v=0\) and \(v=11\text{.}\) This means I can use constant bounds for the new integral. Finally, the Jacobian of the transformation is \(11\text{,}\) so I can set up the new integral (note that the new integral, like the old integral, is improper).
The limit does not converge, but diverges to \(-\infty\text{,}\) so I conclude that the integral does not represent a finite volume.
Activity 3.3.6.
Calculate the integral using a change of variables. (There is a transformation which will simplify both the region of integration and the integrand.)
The region of integration \(D\) is the area bounded above by \(y = x^2 + 1\text{,}\) below by \(y = x^2 +2\text{,}\) on the left by the line \(x=2\text{,}\) and on the first by the line \(x=4\text{.}\) This region is shown in Figure 3.3.2.
It's not entirely clear what is happening with the integrand, so I'll work on making eaiser bounds. The \(x=2\) and \(x=4\) lines are fine, so I will set \(u=x\text{.}\) The other lines can be re-arranged as \(y-x^2 = 1\) and \(y-x^2 = 2\text{,}\) so I'll set \(v = y-x^2\text{,}\) so that the new bounds are \(v=1\) and \(v=2\text{.}\) This gives me a region with constant bounds; let me see what happens to the integrand. I can invert the transformation with \(x = u\) and \(y = v+u^2\text{.}\) The integrand becomes \(\frac{u^3 + 4u + u}{-v} = -\frac{u^3 + 4u + u}{v}\text{,}\) which is much more reasonable. Finally, the Jacobian of this tranformation is \(1\text{.}\)
Activity 3.3.7.
Calculate the integral using a change of variables. (There is a transformation which will simplify both the region of integration and the integrand.)
The region of integration \(D\) is the diamon with coordinates \((4,0)\text{,}\) \((10,6)\text{,}\) \((10,-6)\) and \((16,0)\text{,}\) as show in Figure 3.3.3.
It's not entirely clear what is happening with the integrand, so I'll work on making easier bounds. The lines that bound the diamond are \(x+y = 4\text{,}\) \(x+y = 16\text{,}\) \(x-y = 4\text{,}\) and \(x-y = 16\text{.}\) I could try \(u = x+y\) and \(v = x-y\text{.}\) The integrand factors into \(\sqrt{(x+y)(x-y)}\text{,}\) which would translate into \(\sqrt{uv}\text{,}\) which would work. If I invert, we get \(x = \frac{u+v}{2}\) and \(y = \frac{u-v}{2}\text{.}\) The Jacobian of this transformation is \(\frac{1}{2}\text{.}\) In the calculation, after I seperate the integrals, I notice that are exactly the same. Therefore, I just do one of the integral and square the result.
Subsection 3.3.3 Activity
Activity 3.3.8.
Use polar coordinates to solve this integral of the function \(f(x,y) = \sqrt{x^2 + y^2}\) on the wedge of the circle of radius \(3\) between \(\theta = \frac{3\pi}{4}\) and \(\theta = \frac{5\pi}{4}\text{.}\)
Since \(r = \sqrt{x^2+y^2}\text{,}\) the integrand here is simply \(r\text{.}\) The region is a section of a circle, so the radius will vary from \(0\) to \(3\) and the angle between the two angles given. The Jacobian for polar coordinates is \(r\text{.}\)
Activity 3.3.9.
Use polar coordinates to solve this integral of the function \(f(x,y) = x^2 - y^2\) on the quarter circle of radius \(4\) which is above line \(y=-x\text{,}\) below the line \(y=x\) and include the positive \(x\) axis.
The region is a section of a circle. The radius will range from \(0\) to \(4\) and, for this section, I can take \(\theta \in \left[ \frac{-\pi}{4}, \frac{\pi}{4} \right]\text{.}\) For the integrand, I replace \(x = r \cos \theta\) and \(y = r \sin \theta\) to get \(r^2(\cos^2 \theta - \sin^2 \theta)\text{.}\) Using trig identities, I can write this as \(r^2 \cos 2\theta\text{.}\) The Jacobian of polar coordinates is \(r\text{.}\)
Activity 3.3.10.
Use polar coordinates to solve this integral of the function \(f(x,y) = (x^2 + y^2)^{\frac{3}{2}}\) on the unit circle centred at \((0,1)\text{.}\)
The region is a an offset circle, so I need its in polar locus. In cartesian coordinates, the circle is \(x^2 + (y-1)^2 = 1\text{.}\) If I replace the cartesian coordinates with their polar replacements and simplifity, I get these calculations.
I can take \(r \in [0, 2 \sin \theta]\text{,}\) since the circle touches the origin at one point. The circle is in the half-plane where \(y\) is positive, so the appropriate range of the angle is \(\theta \in [0, \pi]\text{.}\) If I use \(\theta\) as the outside integral, this gives me a way to setup an interated integral with non-constant bounds in polar coordinates. The integral simplifies to \(r^3\text{.}\) (A asked a computer for the antiderivative of \(\sin^5 \theta\text{.}\))
Activity 3.3.11.
Use polar coordinates to solve this integral of the function \(f(x,y) = 3x^3 - (y-3)^2\) on circle of radius \(2\) centred at the origin.
The region here is easy: \(r \in [0,2]\) and \(\theta \in [0, 2\pi]\text{.}\) The integrand, however, doesn't look that pleasant after I change to polar coordinates: \(3r^3 \cos^3 \theta - (r \sin \theta - 3)^2\text{.}\) Adding the Jacobian gives the new integral. I split it up into several pieces to complete the integral. I did some of the trig integrals using a computer algebra system (though you can see, directly, that some evaluate to zero using symmetry arguments.)
Subsection 3.3.4 Conceptual Review Questions
What are polar coordinates?
How is it that change of variable can be arranged to help either the integrand or the region (and sometimes both)?
What is a Jacobian and what does it measure?