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 views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 15.3 : Double Integrals over General Regions
- Evaluate\( \displaystyle \iint\limits_{D}{{42{y^2} - 12x\,dA}}\) where \(D = \left\{ {\left( {x,y} \right)|0 \le x \le 4,{{\left( {x - 2} \right)}^2} \le y \le 6} \right\}\) Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{2y{x^2} + 9{y^3}\,dA}}\) where \(D\) is the region bounded by \( \displaystyle y = \frac{2}{3}x\) and \(y = 2\sqrt x \). Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{10{x^2}{y^3} - 6\,dA}}\) where \(D\) is the region bounded by \(x = - 2{y^2}\) and \(x = {y^3}\). Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{x\left( {y - 1} \right)\,dA}}\) where \(D\) is the region bounded by \(y = 1 - {x^2}\) and \(y = {x^2} - 3\). Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}}\) where \(D\) is the region bounded by \(y = 2\), \( \displaystyle y = \frac{1}{4}{x^2}\) and the \(y\)-axis. Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{\frac{1}{{{y^{\frac{1}{3}}}\left( {{x^3} + 1} \right)}}\,dA}}\) where \(D\) is the region bounded by \(x = - {y^{\frac{1}{3}}}\), \(x = 3\) and the \(x\)-axis. Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{3 - 6xy\,dA}}\) where \(D\) is the region shown below.
Solution
- Evaluate \(\iint\limits_{D}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}}\) where \(D\) is the region shown below.
Solution
- Evaluate \( \displaystyle \iint\limits_{D}{{7{x^2} + 14y\,dA}}\) where \(D\) is the region bounded by \(x = 2{y^2}\) and \(x = 8\) in the order given below.
- Integrate with respect to \(x\) first and then \(y\).
- Integrate with respect to \(y\) first and then \(x\).
For problems 10 & 11 evaluate the given integral by first reversing the order of integration.
- \( \displaystyle \int_{0}^{3}{{\int_{{2x}}^{6}{{\sqrt {{y^2} + 2} \,dy}}\,dx}}\) Solution
- \( \displaystyle \int_{0}^{1}{{\int_{{ - \sqrt y }}^{{{y^{\,2}}}}{{6x - y\,dx}}\,dy}}\) Solution
- Use a double integral to determine the area of the region bounded by \(y = 1 - {x^2}\) and \(y = {x^2} - 3\). Solution
- Use a double integral to determine the volume of the region that is between the \(xy\)‑plane and\(f\left( {x,y} \right) = 2 + \cos \left( {{x^2}} \right)\) and is above the triangle with vertices \(\left( {0,0} \right)\), \(\left( {6,0} \right)\) and \(\left( {6,2} \right)\). Solution
- Use a double integral to determine the volume of the region bounded by \(z = 6 - 5{x^2}\) and the planes \(y = 2x\), \(y = 2\),\(x = 0\) and the \(xy\)-plane. Solution
- Use a double integral to determine the volume of the region formed by the intersection of the two cylinders \({x^2} + {y^2} = 4\) and \({x^2} + {z^2} = 4\). Solution