If the area between the curves y=√x and y=x^3
is rotated completely about the x-axis,find the
volum of the solid made?
Please show step
2 answers
infinite as stated
First step: draw the graphs!
They intersect at (0,0) and (1,1). You want to rotate a small lens-shaped area.
To find the volume, you can do two methods.
Think of the volume as a stack of discs with holes (washers). The holes are there because a slice of area is rotated around the x-axis, but does not touch it. So, with discs of thickness dx, the volume is
v = ∫[0,1] π(R^2-r^2) dx
where R=√x and r=x^3
v = ∫[0,1] π(x-x^6) dx = 5π/14
Or, you can think of the volume as a set of nested cylinders (shells) of thickness dy. Their height is the distance between the two curves x=y^2 and x=∛y.
v = ∫[0,1] 2πrh dy
where r=y and h=∛y-y^2
v = ∫[0,1] 2πy(∛y-y^2) dy = 5π/14
They intersect at (0,0) and (1,1). You want to rotate a small lens-shaped area.
To find the volume, you can do two methods.
Think of the volume as a stack of discs with holes (washers). The holes are there because a slice of area is rotated around the x-axis, but does not touch it. So, with discs of thickness dx, the volume is
v = ∫[0,1] π(R^2-r^2) dx
where R=√x and r=x^3
v = ∫[0,1] π(x-x^6) dx = 5π/14
Or, you can think of the volume as a set of nested cylinders (shells) of thickness dy. Their height is the distance between the two curves x=y^2 and x=∛y.
v = ∫[0,1] 2πrh dy
where r=y and h=∛y-y^2
v = ∫[0,1] 2πy(∛y-y^2) dy = 5π/14
1 answer