# The region under the curve #y=sqrtx# bounded by #0<=x<=4# is rotated about a) the x axis and b) the y axis. How do you sketch the region and find the volumes of the two solids of revolution?

See below

To graph the region:

sketch the curve

Restrict the domain to

use the

Shade in the regione under the curve

Revolve about the

Use disks.

# = pi[x^2/2]_0^4#

# = 8pi# Revolve about the

#y# axisUse shells.

#V = int_0^4 2pisqrtx dx#

# = 2pi[(2x^(3/2))/3]_0^4#

# = (32pi)/3#

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To sketch the region bounded by ( y = \sqrt{x} ) and ( 0 \leq x \leq 4 ), first plot the graph of ( y = \sqrt{x} ) for ( 0 \leq x \leq 4 ). This graph is a half of the parabola opening to the right, starting from the origin and extending to ( x = 4 ).

a) Rotation about the x-axis: Sketch the region and imagine rotating it around the x-axis. This generates a solid with a cylindrical shape, which has a hole in the center.

To find the volume of this solid, integrate the function ( \pi (y^2) ) with respect to ( x ) from ( x = 0 ) to ( x = 4 ). [ V_x = \int_0^4 \pi (\sqrt{x})^2 , dx ]

b) Rotation about the y-axis: Sketch the region and imagine rotating it around the y-axis. This creates a solid resembling a bowl or a dome.

To find the volume of this solid, integrate the function ( \pi (x^2) ) with respect to ( y ) from ( y = 0 ) to ( y = \sqrt{4} = 2 ). [ V_y = \int_0^2 \pi x^2 , dy ]

Solve the integrals to find ( V_x ) and ( V_y ) to determine the volumes of the two solids.

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