MATHEMATICS COLLEGE

4. Find the volume of the given solid bounded by the elliptic paraboloid z = 4 - x^2 - 4y^2, the cylinder x^2 + y^2 = 1 and the plane z = 0.5. Sketch the region of integration and change the order of integration.

Answers

Answer 1

Answer:

Answer:

2.5π units^3

Step-by-step explanation:

Solution:-

- We will evaluate the solid formed by a function defined as an elliptical paraboloid as follows:-

$z = 4 - x^2 -4y^2$

- To sketch the elliptical paraboloid we need to know the two things first is the intersection point on the z-axis and the orientation of the paraboloid ( upward / downward cup ).

- To determine the intersection point on the z-axis. We will substitute the following x = y = 0 into the given function. We get:

$z = 4 - 0 -4*0 = 4$

- The intersection point of surface is z = 4. To determine the orientation of the paraboloid we see the linear term in the equation. The independent coordinates ( x^2 and y^2 ) are non-linear while ( z ) is linear. Hence, the paraboloid is directed along the z-axis.

- To determine the cup upward or downwards we will look at the signs of both non-linear terms ( x^2 and y^2 ). Both non-linear terms are accompanied by the negative sign ( - ). Hence, the surface is cup downwards. The sketch is shown in the attachment.

- Theboundary conditions are expressed in the form of a cylinder and a plane expressed as:

$x^2 + y^2 = 1\n\nz = 4$

- To cylinder is basically an extension of the circle that lies in the ( x - y ) plane out to the missing coordinate direction. Hence, the circle ( x^2 + y^2 = 1 ) of radius = 1 unit is extended along the z - axis ( coordinate missing in the equation ).

- The cylinder bounds the paraboloid in the x-y plane and the plane z = 0 and the intersection coordinate z = 4 of the paraboloid bounds the required solid in the z-direction. ( See the complete sketch in the attachment )

- To determine the volume of solid defined by the elliptical paraboloid bounded by a cylinder and plane we will employ the use of tripple integrals.

- We will first integrate the solid in 3-dimension along the z-direction. With limits: ( z = 0 , $z = 4 - x^2 -4y^2$ ). Then we will integrate the projection of the solid on the x-y plane bounded by a circle ( cylinder ) along the y-direction. With limits: ( $y = - √(1 - x^2)$ , $y = √(1 - x^2)$ ). Finally evaluate along the x-direction represented by a 1-dimensional line with end points ( -1 , 1 ).

- We set up our integral as follows:

$V_s = \int\int\int {} \, dz.dy.dx$

- Integrate with respect to ( dz ) with limits: ( z = 0 , $z = 4 - x^2 -4y^2$ ):

$V_s = \int\int [ {4 - x^2 - 4y^2} ] \, dy.dx$

- Integrate with respect to ( dy ) with limits: ( $y = - √(1 - x^2)$ , $y = √(1 - x^2)$ )

$V_s = \int [ {4y - x^2.y - (4)/(3) y^3} ] \, | .dx\n\nV_s = \int [ {8√(( 1 - x^2 )) - 2x^2*√(( 1 - x^2 )) - (8)/(3) ( 1 - x^2 )^(3)/(2) } ] . dx$

- Integrate with respect to ( dx ) with limits: ( -1 , 1 )

$V_s = [ 4. ( arcsin ( x ) + x√(1 - x^2) ) - (arcsin ( x ) - 2x ( 1 -x^2 )^(3)/(2) + x√(1 - x^2) )/(2) - ( 3*arcsin ( x ) + 2x ( 1 -x^2 )^(3)/(2) + 3x√(1 - x^2) )/(3) ] | \limits^1_-_1\n\nV_s = [ (5)/(2) *arcsin ( x ) + (5)/(3)*x ( 1 -x^2 )^(3)/(2) + (5)/(2) *x√(1 - x^2) ) ] | \limits^1_-_1\n\nV_s = [ (5\pi )/(2) + 0 + 0 ] \n\nV_s = (5\pi )/(2)$

Answer: The volume of the solid bounded by the curves is ( 5π/2 ) units^3.

Answer 2

Answer:

Final answer:

The volume of the bounded region is found by setting up a triple integral, changing to cylindrical coordinates, and integrating to get 3.5π. The region of integration is a solid capped by an elliptic paraboloid, lying inside the unit circle above the xy-plane. Changing the order of integration doesn't apply here as the given order is already the most ideal.

Explanation:

The subject of this question is

Calculating Volume

in integral calculus, specifically dealing with triple integrals. Given the equations z = 4 - x^2 - 4y^2, x^2 + y^2 = 1, and z = 0, we find the volume by setting up a triple integral. In cylindrical coordinates, this is ∫ ∫ (4 - x^2 - 4y^2) rdrdθ from θ=0 to 2π and r=0 to 1. Changing to cylindrical coordinates, x = rcosθ and y = rsinθ, gives ∫ ∫ (4 - r^2) rdrdθ. This evaluates to π(4r - (r^2)/2) evaluated from 0 to 1, which simplifies to π(4 - 0.5) = 3.5π.

Sketching the Region of Integration

, the integrand and bounds describe a solid capped by the elliptic paraboloid and lying above the xy-plane inside the unit circle. The request to 'change the order of integration' would apply if this were an improper triple integral being evaluated in Cartesian coordinates. Here, the order of integration (r, then θ) is itself the most simple and meaningful approach.

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4. Find the volume of the given solid bounded by the elliptic paraboloid z = 4 - x^2 - 4y^2, the cylinder x^2 + y^2 = 1 and the plane z = 0.5. Sketch the region of integration and change the order of integration.

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Final answer:

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