{"id":204,"date":"2021-07-30T17:32:02","date_gmt":"2021-07-30T17:32:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=204"},"modified":"2022-11-01T04:44:07","modified_gmt":"2022-11-01T04:44:07","slug":"putting-it-together-multiple-integration","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/putting-it-together-multiple-integration\/","title":{"raw":"Putting it Together: Multiple Integration","rendered":"Putting it Together: Multiple Integration"},"content":{"raw":"

Finding the Volume of l\u2019Hemisph\u00e8ric<\/h2>\r\nFind the volume of the spherical planetarium in l\u2019Hemisph\u00e8ric in Valencia, Spain, which is five stories tall and has a radius of approximately [latex]50[\/latex] ft,\u00a0using the equation [latex]x^2+y^2+z^2=r^2[\/latex].\r\n\r\n\"A\r\n

Solution<\/h3>\r\nWe calculate the volume of the ball in the first octant, where [latex]x\\ge{0}, y\\ge{0}, \\text{ and } z\\ge{0}[\/latex],\u00a0using spherical coordinates, and then multiply the result by [latex]8[\/latex]\u00a0for symmetry. Since we consider the region [latex]D[\/latex]\u00a0as the first octant in the integral, the ranges of the variables are\r\n

[latex]0\\le{\\varphi}\\le{\\frac{\\pi}{2}},0\\le{\\rho}\\le{r,0}\\le\\theta\\le\\frac{\\pi}{2}[\/latex]Therefore,<\/p>\r\n[latex]\\hspace{5cm}\\large{\\begin{align}\r\n\r\nV&=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz = 8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi} \\rho^{2}\\sin\\theta d\\varphi d\\rho d\\theta \\\\\r\n\r\n&=8\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2} \\ d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=r}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\\r\n\r\n&=8\\left(\\frac{\\pi}2\\right)\\left(\\frac{r^3}3\\right)(1) \\\\\r\n\r\n&=\\frac43\\pi{r}^3\r\n\r\n\\end{align}}[\/latex]\r\n\r\nThis exactly matches with what we knew. So for a sphere with a radius of approximately [latex]50[\/latex] ft, the volume is [latex]\\frac{4}{3}\\pi(50)^3\\approx 523,600[\/latex] ft3<\/sup>.","rendered":"

Finding the Volume of l\u2019Hemisph\u00e8ric<\/h2>\n

Find the volume of the spherical planetarium in l\u2019Hemisph\u00e8ric in Valencia, Spain, which is five stories tall and has a radius of approximately [latex]50[\/latex] ft,\u00a0using the equation [latex]x^2+y^2+z^2=r^2[\/latex].<\/p>\n

\"A<\/p>\n

Solution<\/h3>\n

We calculate the volume of the ball in the first octant, where [latex]x\\ge{0}, y\\ge{0}, \\text{ and } z\\ge{0}[\/latex],\u00a0using spherical coordinates, and then multiply the result by [latex]8[\/latex]\u00a0for symmetry. Since we consider the region [latex]D[\/latex]\u00a0as the first octant in the integral, the ranges of the variables are<\/p>\n

[latex]0\\le{\\varphi}\\le{\\frac{\\pi}{2}},0\\le{\\rho}\\le{r,0}\\le\\theta\\le\\frac{\\pi}{2}[\/latex]Therefore,<\/p>\n

[latex]\\hspace{5cm}\\large{\\begin{align} V&=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz = 8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi} \\rho^{2}\\sin\\theta d\\varphi d\\rho d\\theta \\\\ &=8\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2} \\ d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=r}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\ &=8\\left(\\frac{\\pi}2\\right)\\left(\\frac{r^3}3\\right)(1) \\\\ &=\\frac43\\pi{r}^3 \\end{align}}[\/latex]<\/p>\n

This exactly matches with what we knew. So for a sphere with a radius of approximately [latex]50[\/latex] ft, the volume is [latex]\\frac{4}{3}\\pi(50)^3\\approx 523,600[\/latex] ft3<\/sup>.<\/p>\n\n\t\t\t

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Candela Citations<\/h3>\n\t\t\t\t\t
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