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Find the volume of the region bounded by the elliptic paraboloid z = 4 – x^2...

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teun | Student, College Freshman | eNotes Newbie

Posted September 11, 2013 at 10:19 AM via web

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Find the volume of the region bounded by the elliptic paraboloid z = 4 – x^2 –1/4y^2 and the plane z = 0?

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freemihai | College Teacher | (Level 1) Adjunct Educator

Posted September 11, 2013 at 10:56 AM (Answer #1)

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Put z=0 in elliptic paraboloid equation:

`4 – x^2 –(1/4)y^2 = 0`  ( i have assumed that `1/4`  is coefficient)

`16 - 4x^2 - y^2 = 0`

Find limits of integration letting first y = 0:

`16 - 4x^2 = 0 => 4x^2 = 16`

Divide by 4:

`x^2 = 16/4`

Take square roots to the left and right sides:

`sqrt x^2 = sqrt 4`

Calculate and get only the positive result:

`x = 2`

Find limits of integration for `y^2 = 16 - 4x^2` . Take square roots:

`y = sqrt(16 - 4x^2) => y = sqrt(4(4 - x^2))`

`y = 2sqrt(4 - x^2)`

Calculate volume of first octant of elliptic paraboloid:

`V = int_0^2 int_0^(2sqrt(4 - x^2)) (4 – x^2 –(1/4)y^2) dy dx`

`V = int_0^2 (4y - x^2*y - y^3/12)|_0^(2sqrt(4 - x^2))`

`V = int_0^2 (8sqrt(4 - x^2) - 2x^2sqrt(4 - x^2) - (2/3)sqrt(4 - x^2)^3)dx`

`V = int_0^2 ((16/3)sqrt(4 - x^2) - (4/3)x^2sqrt(4 - x^2))dx`

`V = int_0^2 ((16/3)sqrt(4 - x^2))dx - int_0^2 ((4/3)x^2sqrt(4 - x^2))dx`

Calculate each integral separately:

`I1 = int_0^2 ((16/3)sqrt(4 - x^2))dx`

We can work with trigonometric substitutions:

`x/2 = sin u => x^2/4 = sin^2 u`

`(dx)/2 = cos u du`

`I1 = int_0^(pi/2) (16/3)sqrt(1 - sin^2u)*(2cos u)du`

`I1 = (32/3) int_0^(pi/2) cos u*sqrt(cos^2 u) du`

`I1 = (32/3) int_0^(pi/2) cos u*cos udu` 

`I1 = (32/3) int_0^(pi/2) cos^2 u du`

`I1 = (32/3) int_0^(pi/2) (1/2)*(cos2u + 1)du`

`I1 = (32/6) int_0^(pi/2) cos 2u du + (32/6) int_0^(pi/2) du`

`I1 = 32/12*(sin 2u)|_0^(pi/2) + (32/6)(pi/2 - 0)` `I1 = 32/12*(sin pi - sin 0) + 16pi/6`

`I1 = 8pi/3`

`I2 = int_0^2 ((4/3)x^2sqrt(4 - x^2))dx`

We can work with trigonometric substitutions:

`x/2 = sin u => x^2/4 = sin^2 u`

`(dx)/2 = cos u du` `I2 = int_0^(pi/2) ((4/3)4sin^2 u*sqrt(cos^2 u)*(2cos u)du` `I2 = (8/3)int_0^(pi/2) 4sin^2 u*cos^2 u du `

You know that `2sin u*cos u = sin 2u` , so `4sin^2 u*cos^2 u = sin^2 2u` :

`I2 = (8/3)int_0^(pi/2) sin^2 2u du`

I2`= (8/3)int_0^(pi/2) (1/2)(1 - cos 4u) du`

`I2 = (4/3)int_0^(pi/2)du - (4/3)int_0^(pi/2) cos 4u du`

`I2 = (4/3)(pi/2 - 0) - (1/3) (sin (4pi/2) - sin 0)`

`I2 = 2pi/3`

`V = I1 - I2`

`V = 8pi/3 - 2pi/3 = 2pi`

We just have calculated the volume of first octant of elliptic paraboloid, so the for the entire volume, multiply by 4:

`V = 8pi`

Answer: `V = 8pi`

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