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If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
Section 15.7 : Triple Integrals in Spherical Coordinates
- Evaluate \( \displaystyle \iiint\limits_{E}{{4{y^2}\,dV}}\) where \(E\) is the sphere \({x^2} + {y^2} + {z^2} = 9\).
- Evaluate \( \displaystyle \iiint\limits_{E}{{3x - 2y\,dV}}\) where \(E\) is the region between the spheres \({x^2} + {y^2} + {z^2} = 1\) and \({x^2} + {y^2} + {z^2} = 4\) with \(z \le 0\).
- Evaluate \( \displaystyle \iiint\limits_{E}{{2yz\,dV}}\) where \(E\) is the region inside both \({x^2} + {y^2} + {z^2} = 16\) and \(z = \sqrt {3{x^2} + 3{y^2}} \) that is in the 1st octant.
- Evaluate \( \displaystyle \iiint\limits_{E}{{{z^2}\,dV}}\) where \(E\) is the region between the spheres \({x^2} + {y^2} + {z^2} = 4\) and \({x^2} + {y^2} + {z^2} = 25\) and inside \(\displaystyle z = - \sqrt {\frac{1}{3}{x^2} + \frac{1}{3}{y^2}} \).
- Evaluate \( \displaystyle \iiint\limits_{E}{{5{y^2}\,dV}}\) where \(E\) is the portion of \({x^2} + {y^2} + {z^2} = 4\) with \(x \le 0\).
- Evaluate \( \displaystyle \iiint\limits_{E}{{2 + 16x\,dV}}\) where \(E\) is the region between the spheres \({x^2} + {y^2} + {z^2} = 1\) and \({x^2} + {y^2} + {z^2} = 4\) with \(y \ge 0\) and \(z \le 0\).
- Evaluate the following integral by first converting to an integral in cylindrical coordinates. \[\int_{0}^{2}{{\int_{{ - \sqrt {4 - {x^{\,2}}} }}^{0}{{\int_{{\sqrt {5{x^{\,2}} + 5{y^{\,2}}} }}^{{\sqrt {9 - {x^{\,2}} - {y^{\,2}}} }}{{\,\,\,7x\,\,\,dz}}\,dy}}\,dx}}\]
- Evaluate the following integral by first converting to an integral in cylindrical coordinates. \[\int_{{ - \sqrt 5 }}^{{\sqrt 5 }}{{\int_{0}^{{\sqrt {5 - {y^{\,2}}} }}{{\int_{{ - \sqrt {10 - {x^{\,2}} - {y^{\,2}}} }}^{{ - \sqrt {{x^{\,2}} + {y^{\,2}}} }}{{\,\,\,3x{z^2}\,\,\,dz}}\,dx}}\,dy}}\]
- Use a triple integral in spherical coordinates to derive the volume of a sphere with radius \(a\).