A, B and C did a test, A got 3/5 of
1 answer:
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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:-
- 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:
- 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.
- The boundary conditions are expressed in the form of a cylinder and a plane expressed as:
- 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 , ). 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: (
,
). Finally evaluate along the x-direction represented by a 1-dimensional line with end points ( -1 , 1 ).
- We set up our integral as follows:
- Integrate with respect to ( dz ) with limits: ( z = 0 , ):
- Integrate with respect to ( dy ) with limits: ( ,
)
- Integrate with respect to ( dx ) with limits: ( -1 , 1 )
Answer: The volume of the solid bounded by the curves is ( 5π/2 ) units^3.