Course Notes for Calculus IV

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.

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.

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.

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.

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.

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 \(1\text{.}\)

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.

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 \(r\text{.}\)

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 \(r\text{.}\)

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 \(\sin^5 \theta\text{.}\))

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.)