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The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any met...Asked by Ron
The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method.
x = (y − 9)^2, x = 16; about y = 5
x = (y − 9)^2, x = 16; about y = 5
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Answered by
mathhelper
Assuming you made a decent sketch,
x = (y-9)^2
± √x = y - 9
y = 9 + √x and y = 9 - √x
notice the parabola has vertex at (0,9) and it intersects x = 16
and y = 5 at (16,5)
upper height above y = 5 is y = 9+√x - 5 = 4 + √x <---- larger radius
lower height above y = 5 is y = 9-√x - 5 = 4 - √x <----- smaller radius
volume = π∫ ( (4+√x)^2 - (4 - √x)^2 ) dx from 0 to 16
= π ∫ 16√x dx from 0 to 16
= π [ 32/3 x^(3/2) ] from 0 to 16
= π ( (32/3)(64) - 0 )
= 2048/3 π
better check my steps, my 82 year old brain ain't what it used to be.
x = (y-9)^2
± √x = y - 9
y = 9 + √x and y = 9 - √x
notice the parabola has vertex at (0,9) and it intersects x = 16
and y = 5 at (16,5)
upper height above y = 5 is y = 9+√x - 5 = 4 + √x <---- larger radius
lower height above y = 5 is y = 9-√x - 5 = 4 - √x <----- smaller radius
volume = π∫ ( (4+√x)^2 - (4 - √x)^2 ) dx from 0 to 16
= π ∫ 16√x dx from 0 to 16
= π [ 32/3 x^(3/2) ] from 0 to 16
= π ( (32/3)(64) - 0 )
= 2048/3 π
better check my steps, my 82 year old brain ain't what it used to be.