Volume of a Solid of Revolution?Find the volume if the region enclosing y = x 3 , x = 0 ,, and y = 8 is rotated about the given line. So the axis of rotation that was given was the y axis, but I'm a bit confused on what the given bounds of y = 8 and x = 0 have to do with it, but I pretty much ignored them and put the given equation in terms of y and used the integral ∫ 0 7 ( 2 2 − ( y 3 ) 2 ) to solve, is this at all right? Any help would be appreciated Also looking for some guidance on how I'd go about finding the volume if the axis of rotation was something like x = 6.

Question

Volume of a Solid of Revolution?Find the volume if the region enclosing   y  =      x    3    ,  x  =  0  ,, and   y  =  8 is rotated about the given line. So the axis of rotation that was given was the y axis, but I&#039;m a bit confused on what the given bounds of   y  =  8 and   x  =  0 have to do with it, but I pretty much ignored them and put the given equation in terms of y and used the integral       ∫    0    7    (      2    2    −  (      y    3        )    2    ) to solve, is this at all right? Any help would be appreciated Also looking for some guidance on how I&#039;d go about finding the volume if the axis of rotation was something like   x  =  6.

Rihanna Robles · Accepted Answer

Step 1Note that   y  =      x    3    ,    x  =  0 and   y  =  8 only specify the area to be rotated. The volume is then defined by rotating this area around a given axis.Step 2For instance, if the axis of rotation is   x  =  6, the volume can be integrated with the disk method       ∫    0    8    π  [      6    2    −  (  6  −      y    3        )    2    ]  d  y

Glenn Hopkins · Accepted Answer

Step 1Now, we can go about finding the volume. Since we are revolving around the y-axis, we can use the following formula:   V  =      ∫    a    b    A  (  y  )  d  yAnd, we know that   A  (  y  )  =  π  (  (  outer radius      )    2    −  (  inner radius      )    2    ), where the radius is just the distance from the curve to the axis of rotation.Since the axis of rotation is the y-axis (equivalently,   x  =  0), we know that are distances must be x-distances, so we must solve our y-functions in terms of x.Our outer radius is just the distance from the curve   y  =      x    3   to the y-axis, we which is given by       y          1              /            3        −  0  =      y          1              /            3      .Step 2Our inner radius is the distance from the y-axis to itself, which is 0.Therefore, the expression for the area in terms of y is:  A  (  y  )  =  π  (  (      y          1              /            3            )    2    −  (  0      )    2    )  =  π  (      y          2              /            3        )Now, we just need to plug back into the equation for the volume, and figure out the bounds. Since this is a dy integral, we need to know the y-bound. From the region, we can see that the integral should go from 0 to 8, since the region is bounded by   y  =  0 and   y  =  8Therefore, our integral becomes:   V  =      ∫    0    8    π  (      y          2              /            3        )  d  yIf the axis of rotation changes, you just need to change your expression for the area. For example, for an axis of rotation of   x  =  6, you can keep the area in terms of y, but now your inner and outer radii expressions will be different.

Find the volume if the region enclosing y=x^3, x=0, and y=8 is rotated about the given line.

Answered question

Answer & Explanation

New Questions in High school geometry