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Math 233 Name: Open with Kami Quiz Question 1. Evaluate the triple integral int_(-a)^(a)int_(-sqrt(a^(2)-x^(2)))^(sqrt(a^(2)-x^(2)))int_(z=0)^(z=h)dzdydx whère a and h are arbitrary constants.

Math 233233\newlineName:\newlineOpen with Kami\newlineQuiz\newlineQuestion 11. Evaluate the triple integral\newline∫−aa∫−a2−x2a2−x2∫z=0z=hdzdydx \int_{-a}^{a} \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}} \int_{z=0}^{z=h} d z d y d x \newlinewhère a a and h h are arbitrary constants.

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Q. Math 233233\newlineName:\newlineOpen with Kami\newlineQuiz\newlineQuestion 11. Evaluate the triple integral\newline∫−aa∫−a2−x2a2−x2∫z=0z=hdzdydx \int_{-a}^{a} \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}} \int_{z=0}^{z=h} d z d y d x \newlinewhère a a and h h are arbitrary constants.
  1. Integrate with respect to zz: First, integrate with respect to zz from 00 to hh.\newline∫z=0z=hdz=z∣z=0z=h=h−0=h\int_{z=0}^{z=h} dz = z \bigg|_{z=0}^{z=h} = h - 0 = h
  2. Integrate with respect to y: Now, integrate the result with respect to yy from −a2−x2-\sqrt{a^2 - x^2} to a2−x2\sqrt{a^2 - x^2}.∫−a2−x2a2−x2h dy=h⋅y∣−a2−x2a2−x2=h⋅[a2−x2−(−a2−x2)]=2ha2−x2\int_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} h \, dy = h \cdot y \bigg|_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} = h \cdot [\sqrt{a^2 - x^2} - (-\sqrt{a^2 - x^2})] = 2h\sqrt{a^2 - x^2}
  3. Integrate with respect to xx: Finally, integrate the result with respect to xx from −a-a to aa.∫−aa2ha2−x2 dx\int_{-a}^{a} 2h\sqrt{a^2 - x^2} \, dx
  4. Recognize the integral: Recognize that the integral represents the area of a semicircle with radius aa, multiplied by the height hh.\newlineArea of a semicircle = (1/2)πa2(1/2)\pi a^2\newlineSo, ∫−aa2ha2−x2dx=2h×(1/2)πa2=πa2h\int_{-a}^{a} 2h\sqrt{a^2 - x^2} dx = 2h \times (1/2)\pi a^2 = \pi a^2 h

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