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
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Section 15.5 : Triple Integrals
- Evaluate \( \displaystyle \int_{2}^{3}{{\int_{{ - 1}}^{4}{{\int_{1}^{0}{{4{x^2}y - {z^3}\,dz}}\,dy}}\,dx}}\) Solution
- Evaluate \( \displaystyle \int_{0}^{1}{{\int_{0}^{{{z^{\,2}}}}{{\int_{0}^{3}{{y\cos \left( {{z^5}} \right)\,dx}}\,dy}}\,dz}}\) Solution
- Evaluate \( \displaystyle \iiint\limits_{E}{{6{z^2}\,dV}}\) where \(E\) is the region below \(4x + y + 2z = 10\) in the first octant. Solution
- Evaluate \( \displaystyle \iiint\limits_{E}{{3 - 4x\,dV}}\) where \(E\) is the region below \(z = 4 - xy\) and above the region in the \(xy\)-plane defined by \(0 \le x \le 2\), \(0 \le y \le 1\). Solution
- Evaluate \( \displaystyle \iiint\limits_{E}{{12y - 8x\,dV}}\) where \(E\) is the region behind \(y = 10 - 2z\) and in front of the region in the \(xz\)-plane bounded by \(z = 2x\), \(z = 5\) and \(x = 0\). Solution
- Evaluate \( \displaystyle \iiint\limits_{E}{{yz\,dV}}\) where \(E\) is the region bounded by \(x = 2{y^2} + 2{z^2} - 5\) and the plane \(x = 1\). Solution
- Evaluate \( \displaystyle \iiint\limits_{E}{{15z\,dV}}\) where \(E\) is the region between \(2x + y + z = 4\) and \(4x + 4y + 2z = 20\) that is in front of the region in the \(yz\)-plane bounded by \(z = 2{y^2}\) and \(z = \sqrt {4y} \). Solution
- Use a triple integral to determine the volume of the region below \(z = 4 - xy\) and above the region in the \(xy\)-plane defined by \(0 \le x \le 2\), \(0 \le y \le 1\). Solution
- Use a triple integral to determine the volume of the region that is below \(z = 8 - {x^2} - {y^2}\) above \(z = - \sqrt {4{x^2} + 4{y^2}} \) and inside \({x^2} + {y^2} = 4\). Solution