Help with finding the volume of a solid of revolutionRevolve x 2 + 4 y 2 = 4.a. About y = 2.b. About x = 2.The answer at the back of the book is the same for both a . 2 π ∫ 0 2 8 1 − x 2 4 d x = 78.95684 b . 2 π ∫ 0 1 8 4 − 4 y 2 d y = 78.95684I tried splitting the ellipse into its upper and lower half but I still couldn't get the answer. I get that its symmetrical that's why it's multiplied to two but I don't know how you get the equation inside the integral.

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Help with finding the volume of a solid of revolutionRevolve       x    2    +  4      y    2    =  4.a. About   y  =  2.b. About   x  =  2.The answer at the back of the book is the same for both  a  .  2  π      ∫          0              2        8      1    −                  x                  2                    4        d  x  =  78.95684  b  .  2  π      ∫          0              1        8      4    −    4                  y                  2                      d  y  =  78.95684I tried splitting the ellipse into its upper and lower half but I still couldn&#039;t get the answer. I get that its symmetrical that&#039;s why it&#039;s multiplied to two but I don&#039;t know how you get the equation inside the integral.

Darren Maxwell · Accepted Answer

Step 1Let&#039;s try the shell method instead of the disk method? Let&#039;s consider the problem b).The distance from the axis of rotation is   (  2  −  x  ). Considering the top half of the ellipse,  A  =  2  π      ∫          −      2        2    (  2  −  x  )      1    −                  x        2            4               d    x  =  4      π    2  Step 2Doubling this gives us the volume of the whole ellipsoid,   8      π    2    ≈  78.95.I will think about the disk method and maybe add it on.

Leyla Bishop · Accepted Answer

Step 1About   y  =  2: It seems natural to take cross-sections perpendicular to the x-axis. A cross-section at x is a washer, with outer radius   R  (  x  )  =  2  +      1    −                  x        2            4       and inner radius   r  (  x  )  =  2  −      1    −                  x        2            4      .The volume is       ∫          −      2        2    (  π      R    2    (  x  )  −  π      r    2    (  x  )  )    d  x  .When we expand and simplify, there is a lot of cancellation, just like in the first problem, and we get       ∫          −      2        2    8  π      1    −                  x        2            4          d  x  .It is a good idea to take advantage of symmetry. So we integrate from 0 to 2, and multiply the result by 2.About   x  =  2: Cylindrical shells seem like a good idea, but we can also do it by taking slices perpendicular to the y-axis.Step 2Again we get a washer. The outer radius is the distance from   x  =  −      4    −    4          y      2       to   x  =  2, and the inner radius is the distance from       4    −    4          y      2       to 2. When we set up the calculation, we get a lot of cancellation, and the result is      ∫          y      =      −      1        1    16  π      1    −          y      2          d  y  .Again, we can take advantage of symmetry, integrate from 0 to 1 and double the result.The answers happen to be the same. We can see this by making the substitution   x  =  2  t in the first integral.

Revolve x^2+4y^2=4. a) About y=2. b) About x=2.

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