Finding the volume of a sphere with a triple integral and trig subHow is trigonometric substitution done with a triple integral? For instance, 8 ∫ 0 r ∫ 0 r 2 − x 2 ∫ 0 r 2 − x 2 − y 2 ( 1 ) d z d y d xHere the limits have been chosen to slice an 8th of a sphere through the origin of radius r, and to multiply this volume by 8. Without converting coordinates, how might a trig substitution be done to solve this?

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Finding the volume of a sphere with a triple integral and trig subHow is trigonometric substitution done with a triple integral? For instance,   8      ∫    0    r        ∫    0                            r          2                −                  x          2                          ∫    0                            r          2                −                  x          2                −                  y          2                      (  1  )  d  z  d  y  d  xHere the limits have been chosen to slice an 8th of a sphere through the origin of radius r, and to multiply this volume by 8. Without converting coordinates, how might a trig substitution be done to solve this?

Dillan Brock · Accepted Answer

Step 1The volume of the sphere   B  (  0  ,  r  )  =  {  (  x  ,  y  ,  z  )  :      x    2    +      y    2    +      z    2    ≤      r    2    } is usually calculated as follows: Make the change of variable   x  =  r  cos  ⁡  θ  sin  ⁡  ϕ  ;     y  =  r  sin  ⁡  θ  sin  ⁡  ϕ  ;     z  =  r  cos  ⁡  ϕ, with the Jacobian equal to       r    2    sin  ⁡  ϕ.Step 2  V  =      ∫          B      (      0      ,      r      )        1  d  x  =      ∫    0    r        ∫    0          2      π            ∫    0    π        r    2    sin  ⁡  ϕ  d  ϕ  d  θ  d  r  =            4      π              r        3              3

saillantpq · Accepted Answer

Step 1The innermost integral has the value             r      2        −          x      2        −          y      2      .The next integral we are faced with is   I  (  x  )  :=      ∫    0                            r          2                −                  x          2                                r      2        −          x      2        −          y      2           d  y     .During the integration x is constant. &quot;Trigonometric substitution&quot; means here that we somehow should use   1  −      sin    2    ⁡  t  ≡      cos    2    ⁡  t to get rid of the square root. Therefore we substitute  y  :=            r      2        −          x      2        sin  ⁡  t     ,    d  y  =            r      2        −          x      2        cos  ⁡  t     d  t    (  0  ≤  t  ≤            π      2        )     ,and I(x) becomes  I  (  x  )  =  (      r    2    −      x    2    )         ∫    0          π              /            2            cos    2    ⁡  t     d  t     .Step 2Now use the rule       cos    2    ⁡  (  ω  t  ) or       sin    2    ⁡  (  ω  t  ) integrated over an integer number of quarter periods gives half of the length of the integration interval&quot; and obtain   I  (  x  )  =            π      4        (      r    2    −      x    2    ).It remains to compute the outermost integral:      v    o    l    (      B    r    )  =  8      ∫    0    r    I  (  x  )     d  x  =  2  π         ∫    0    r    (      r    2    −      x    2    )     d  x  =  2  π  (      r    2    x  −                    x        3            3        )                    |              0    r    =                    4        π            3               r    3       .

How is trigonometric substitution done with a triple integral? For instance, 8 int_0^r int_0^{sqrt{r^2-x^2}} int_0^{sqrt{r^2-x^2-y^2}}(1)dzdydx

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