volume using cross-sections

Calculus
Tutor: None Selected Time limit: 1 Day

find the volume of the solid generated by revolving the region bounded by the given curve and line about the x-axis y = 2square root of x, y =2, x=0

May 15th, 2015

This is the volume defined by a revolving triangular shaped region bounded on the left by the y axis (x=0), from above by y=2, below and on the side by the curve y = sqrt(x).

We first compute the volume of the cylinder defined by the revolving region bounded by y=2 , x=0 and x=4 (the intersection of y=2 and y = sqrt(x)).

The volume of this cylinder is simply base area* height =   pi* 2^2 *4=16pi

From this we have to subtract the volume defined by the revolving curve y = sqrt(x) and and the line x= 4

Vol under the curve = pi integral over  y^2 dx from x=0 to 4  = pi integral over  x from 0 to 4 = pi*x^2 /2 from 0 to 4 = pi(4^2/2 - 2^2/2) = 6 pi

The required volume is then: 16pi - 6pi = 10 pi


May 15th, 2015

hi... its y= 2sqrt of x

May 15th, 2015

Ok - this changes things. I'll give you the corrected answer in a few minutes

 

May 15th, 2015

ok thanks

May 15th, 2015

See the pdf attached above.

May 15th, 2015

volume = pi int from 0 to 1 (2 sqrt(x))^2 dx  = pi int (4 x) = pi (4 x^2/2) from 0 to 1 = 2pi

May 15th, 2015

volume = pi int from 0 to 1 (2 sqrt(x))^2 dx  = pi int (4 x) = pi (4 x^2/2) from 0 to 1 = 2pi

May 15th, 2015

volume = pi int from 0 to 1 (2 sqrt(x))^2 dx  = pi int (4 x) = pi (4 x^2/2) from 0 to 1 = 2pi

May 15th, 2015

it tells me is not correct ;/

May 16th, 2015

volume under 2 sqrt(x) from 0 to 1.pdf

Correction: see the attached

it is 2*pi

May 16th, 2015

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