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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
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Section 15.4 : Double Integrals in Polar Coordinates
- Evaluate \( \displaystyle \iint\limits_{D}{{{y^2} + 3x\,dA}}\) where \(D\) is the region in the 3rd quadrant between \({x^2} + {y^2} = 1\) and \({x^2} + {y^2} = 9\). Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{\sqrt {1 + 4{x^2} + 4{y^2}} \,dA}}\) where \(D\) is the bottom half of \({x^2} + {y^2} = 16\). Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{4xy - 7\,dA}}\) where \(D\) is the portion of \({x^2} + {y^2} = 2\) in the 1st quadrant. Solution
- Use a double integral to determine the area of the region that is inside \(r = 4 + 2\sin \theta \) and outside \(r = 3 - \sin \theta \). Solution
- Evaluate the following integral by first converting to an integral in polar coordinates. \[\int_{0}^{3}{{\int_{{ - \sqrt {9 - {x^{\,2}}} }}^{0}{{\,\,\,{{\bf{e}}^{{x^{\,2}} + {y^{\,2}}}}\,dy}}\,dx}}\] Solution
- Use a double integral to determine the volume of the solid that is inside the cylinder \({x^2} + {y^2} = 16\), below \(z = 2{x^2} + 2{y^2}\) and above the \(xy\)-plane. Solution
- Use a double integral to determine the volume of the solid that is bounded by \(z = 8 - {x^2} - {y^2}\) and \(z = 3{x^2} + 3{y^2} - 4\). Solution