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Some Integration Q's (1 Viewer)

nrlwinner

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Hi. I'm having trouble on some integration. These are from the 2 U Fitzpatrick Book.

12. A hemispherical bowl of radius a units is filled with water to a depth of a/2 units. Find, by integration, the volume of the water.

18. The region bounded by the curve x^1/2 + y^1/2 = 2 and the coordinate axes is rotated about the x-axis. Find the volume of the solid generated.
 

GUSSSSSSSSSSSSS

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Hi. I'm having trouble on some integration. These are from the 2 U Fitzpatrick Book.

12. A hemispherical bowl of radius a units is filled with water to a depth of a/2 units. Find, by integration, the volume of the water.

18. The region bounded by the curve x^1/2 + y^1/2 = 2 and the coordinate axes is rotated about the x-axis. Find the volume of the solid generated.
then post them in the 2u section of this forum, not the mathematics extension II

for the first one: assume you have a circle in the co-ordinate plane, with centre (0,0) and radius = "a" units
therefore equation of circle: x^2 + y^2 = a^2

let y = -a/2 and solve to find the two values of x

then rearrange the equation to get: y^2 = a^2 - x^2
and find the volume when the are between the positive x value and 0 is rotated about the y axis
it will be the pi multiplied by the integral between the positive x value and 0 of a^2 - x^2 dx ......(standard formula for calculating volumes)



ima sorry but i cbb doin the 2nd one, once again u just use the standard formula for finding volumes
 

addikaye03

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Hi. I'm having trouble on some integration. These are from the 2 U Fitzpatrick Book.

12. A hemispherical bowl of radius a units is filled with water to a depth of a/2 units. Find, by integration, the volume of the water.

18. The region bounded by the curve x^1/2 + y^1/2 = 2 and the coordinate axes is rotated about the x-axis. Find the volume of the solid generated.
y^1/2 = 2-x^1/2

y^2=(2-x^1/2)^4= (2)^4+4(2)^3(-x^1/2)+6(2)^2(-x^1/2)^2+4(2)(-x^1/2)^3+(-x^1/2)^4

=16-32x^1/2+24x-8x^3/2+x^2

V=pi int(b-->a) y^2 dx , should be right from there
 

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