General Notice
I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 15.8 : Change of Variables
For problems 1 – 3 compute the Jacobian of each transformation.
- \(x = 4u - 3{v^2}\hspace{0.25in}y = {u^2} - 6v\) Solution
- \(x = {u^2}{v^3}\hspace{0.25in}y = 4 - 2\sqrt u \) Solution
- \(\displaystyle x = \frac{v}{u}\hspace{0.25in}\hspace{0.25in}y = {u^2} - 4{v^2}\) Solution
- If \(R\) is the region inside \(\displaystyle \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{36}} = 1\) determine the region we would get applying the transformation \(x = 2u\), \(y = 6v\) to \(R\). Solution
- If \(R\) is the parallelogram with vertices \(\left( {1,0} \right)\), \(\left( {4,3} \right)\), \(\left( {1,6} \right)\) and \(\left( { - 2,3} \right)\) determine the region we would get applying the transformation \(\displaystyle x = \frac{1}{2}\left( {v - u} \right)\), \(\displaystyle y = \frac{1}{2}\left( {v + u} \right)\) to \(R\). Solution
- If \(R\) is the region bounded by \(xy = 1\), \(xy = 3\), \(y = 2\) and \(y = 6\) determine the region we would get applying the transformation \(\displaystyle x = \frac{v}{{6u}}\), \(y = 2u\) to \(R\). Solution
- Evaluate \(\displaystyle \iint\limits_{R}{{x{y^3}\,dA}}\) where \(R\) is the region bounded by \(xy = 1\), \(xy = 3\), \(y = 2\) and \(y = 6\) using the transformation \(\displaystyle x = \frac{v}{{6u}}\), \(y = 2u\). Solution
- Evaluate \(\displaystyle \iint\limits_{R}{{6x - 3y\,dA}}\) where \(R\) is the parallelogram with vertices \(\left( {2,0} \right)\), \(\left( {5,3} \right)\), \(\left( {6,7} \right)\) and \(\left( {3,4} \right)\) using the transformation \(\displaystyle x = \frac{1}{3}\left( {v - u} \right)\), \(\displaystyle y = \frac{1}{3}\left( {4v - u} \right)\) to \(R\). Solution
- Evaluate \(\displaystyle \iint\limits_{R}{{x + 2y\,dA}}\) where \(R\) is the triangle with vertices \(\left( {0,3} \right)\), \(\left( {4,1} \right)\) and \(\left( {2,6} \right)\) using the transformation \(\displaystyle x = \frac{1}{2}\left( {u - v} \right)\), \(\displaystyle y = \frac{1}{4}\left( {3u + v + 12} \right)\) to \(R\). Solution
- Derive the transformation used in problem 8. Solution
- Derive a transformation that will convert the triangle with vertices \(\left( {1,0} \right)\), \(\left( {6,0} \right)\) and \(\left( {3,8} \right)\) into a right triangle with the right angle occurring at the origin of the \(uv\) system. Solution