In order to be bounded, the domain has to be defined by the inequalities: 6x+2z<=1,z>=7(x^2)+(y^2). Otherwise, it can be checked with RegionPlot3D that domain is not bounded and the volume would be ∞.  Therefore, for these first inequalities the domain is:

a=RegionPlot3D[6x+2z<=1&&z>=7(x^2)+(y^2),{x,-1,1},{y,-1,1},{z,0,5},PlotPoints->60,Boxed->False,AxesOrigin->{0,0,0},AxesLabel->{"x","y","z"}]

If we represent the boundary corresponding to the two last inequalities we have

b=ContourPlot3D[{2x+2y==0,9x-4y==0},{x,-1,1},{y,-1,1},{z,0,5},PlotPoints->60,Boxed->False,AxesOrigin->{0,0,0},AxesLabel->{"x","y","z"}]

With the inequalities we have

RegionPlot3D[2x+2y>=0 && 9x-4y>=0,{x,-5,5},{y,-5,10},{z,0,5},Boxed->False,AxesOrigin->{0,0,0},AxesLabel->{"x","y","z"},PlotPoints->40]

Therefore our domain is the intersection of both which consists of  the portion of the initial graph between both planes in the quarters 4th and 1st

c=Graphics3D[{Arrow[{{1,0,0.2},{0.5,0,0.2}}]}];
Show[{a,b,c}]

The final domain would be the portion of the initial graphic between these to planes.

RegionPlot3D[6x+2z<=1&&z>=7(x^2)+(y^2)&&2x+2y>=0 && 9x-4y>=0,{x,0,0.5},{y,-0.5,0.5},{z,0,5},PlotPoints->100,Boxed->False,AxesOrigin->{0,0,0},AxesLabel->{"x","y","z"}]

If we see it from above we have the part of the plane between the lines 2x+2y=0 (in blue) and 9x-4y=0 (yellow).

a=ContourPlot[{2x+2y==0 , 9x-4y==0},{x,-5,5},{y,-5,15},Axes->True,Frame->False]

If we represent the inequalities, we have:

b=RegionPlot[2x+2y>=0 && 9x-4y>=0,{x,-5,5},{y,-5,15},Axes->True,Frame->False];
Show[{b,a}]

Therefore the horizontal angle θ has to move from the angle of the line 2x+2y=0 (the blue one) which is in the 4th quarter and the angle of the line 9x-4y=0 (yellow) which is in the first quarter.

Therefore to determine the angles we have to consider that the initial one has to be in the 4th quarter and the final one which has to be bigger than the initial one and has to be in the 1st quarter.

Let us start with some computations

z==7(x^2)+(y^2) is an elliptic paraboloid an as it can be written as z=((((x/(1/(Sqrt[7])))))^2)+(y^2), it is centered at the origin ((Subscript[x, 0,])(Subscript[y, 0]))=(0,0) and with semi-axes a=(1/(Sqrt[7])) and b=1. Therefore we apply the transformation:

T={x->(1/(Sqrt[7]))ρ Cos[θ],y->ρ Sin[θ],z->z};

The corresponding Jacobian determinant

Det[D[{((ρ Cos[θ])/(Sqrt[7])),ρ Sin[θ],z},{{ρ,θ,z},1}]]//Simplify

If we apply the transformation to the paraboloid we obtain:

z==7(x^2)+(y^2)/.T//Simplify

For the plane we have:

6x+2z==1/.T

We can compute the limits for variable z from the equation of the paraboloid, so z>=(ρ^2), and the one of the plane so that

Solve[2 z+((6 ρ Cos[θ])/(Sqrt[7]))==1,z]

Therefore the integra for z will be

(∫_(ρ^2)^((1/14) (7-6 (Sqrt[7]) ρ Cos[θ])))\[DifferentialD]z

The radius ρ will move from ρ=0 to the limit given by the intersection of the plane and the paraboloid. We can compute that limit for ρ by means of the system obtained from the equations of both of them.

Solve[{2 z+((6 ρ Cos[θ])/(Sqrt[7]))==1,z==(ρ^2)},{ρ,z}]

As ρ>=0, the only possible solution is the one with + so the integral for ρ is:

(∫_0^((1/14) (-3 (Sqrt[7]) Cos[θ]+(Sqrt[7]) (Sqrt[14+9 ((Cos[θ])^2)]))))\[DifferentialD]ρ

Finally the transformation of the plane 2x+2y=0 is

2x+2y==0/.T

We can simplify dividing by ρ and then we solve to obtain the corresponding initial angle θ. As we said before the initial angle has to be in the 4th quarter

Solve[((2  Cos[θ])/(Sqrt[7]))+2 Sin[θ]==0,θ]

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We take for the initial angle the first solution (with c1=0) -ArcTan[(1/(Sqrt[7]))]=-0.361367 because it is an angle of the 4th quarter (the other solution π-ArcTan[(1/(Sqrt[7]))]=2.78023, belongs to the 2nd quarter)

For the final angle we consider the transformation of the plane

9x-4y==0/.T

Again we divide by ρ and we solve to obtain the angle

Solve[((9  Cos[θ])/(Sqrt[7]))-4  Sin[θ]==0]

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As we deduced before, the final angle hasto be in the 1st quarter and to be bigger than the inicial one so we take the first solution ArcTan[(9/(4 (Sqrt[7])))]=0.704738  which is in the 1st quarter and bigger than the initial one (-0.361367 < 0.704738 ).

Finally, if we insert the Jacobian determinant, the volume can be computed by means of the integral

(∫_(-ArcTan[(1/(Sqrt[7]))])^(ArcTan[(9/(4 (Sqrt[7])))]))(∫_0^((1/14) (-3 (Sqrt[7]) Cos[θ]+(Sqrt[7]) (Sqrt[14+9 ((Cos[θ])^2)]))))(∫_(ρ^2)^((1/14) (7-6 (Sqrt[7]) ρ Cos[θ])))(ρ/(Sqrt[7]))\[DifferentialD]z\[DifferentialD]ρ\[DifferentialD]θ

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RegionPlot3D[6x+3z<=2&&z>=8(x^2)+7(y^2),{x,-1,1},{y,-1,1},{z,0,5},PlotPoints->60,Boxed->False,AxesOrigin->{0,0,0},AxesLabel->{"x","y","z"}]

a=ContourPlot[{x-5y==0 , -5y==0},{x,-5,5},{y,-5,5},Axes->True,Frame->False]

If we represent the inequalities, we have:

b=RegionPlot[x-5y>=0 && -5y>=0,{x,-5,5},{y,-5,5},Axes->True,Frame->False];
Show[{b,a}]

Therefore, the domain is the portion delimited by the angle of the line x-5y=0 in the 3rd quadrant and the angle of line -5y=0 in quadrant 4th/1st.

We have an elliptic paraboloid z=8(x^2)+7(x^2) that can be also written as z=((((x/(1/(Sqrt[8])))))^2)+((((y/(1/(Sqrt[7])))))^2). We thus apply the transformation

T={x->(1/(Sqrt[8]))ρ Cos[θ],y->(1/(Sqrt[7]))ρ Sin[θ],z->z};

The jacobian determinant is

(1/(Sqrt[8]))(1/(Sqrt[7]))ρ

The transformed paraboloid is

z==8(x^2)+7(y^2)/.T//Simplify

And the plane 6x+3z=2,

6x+3z==2/.T

From which we obtein the upper limit for z:

Solve[3 z+((3 ρ Cos[θ])/(Sqrt[2]))==2,z]

In this way, the integral for z is

(∫_(ρ^2)^((1/6) (4-3 (Sqrt[2]) ρ Cos[θ])))\[DifferentialD]z

The limit for ρ is given by the intersection between the plane and the paraboloid.

Solve[{3 z+((3 ρ Cos[θ])/(Sqrt[2]))==2,z==(ρ^2)},{ρ,z}]

We take the positive solution for ρ and the integral will be

(∫_0^((1/12) (-3 (Sqrt[2]) Cos[θ]+(Sqrt[6]) (Sqrt[16+3 ((Cos[θ])^2)]))))\[DifferentialD]ρ

Finally the limit for θ is given by the transforme line

x-5y==0/.T

We simplify ρ and solve for θ

Solve[(( Cos[θ])/(2 (Sqrt[2])))-((5  Sin[θ])/(Sqrt[7]))==0]

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As the initial angle has to be in the secon quadrant we take -π+ArcTan[((Sqrt[(7/2)])/10)]=-2.956647697980422`

For the upper limit for θ we consider the second line

-5y==0/.T

So that

Solve[-((5  Sin[θ])/(Sqrt[7]))==0]

If we take (Subscript[c, 1])=0, we have two solutions θ=0 and θ=π. As the upper limit for θ has to be in the 4th/1st quadrant, we take θ=0. The integral for θ is then

(∫_(-π+ArcTan[((Sqrt[(7/2)])/10)])^0)\[DifferentialD]θ

In this way, the final integral is

(∫_(-π+ArcTan[((Sqrt[(7/2)])/10)])^0)(∫_0^((1/12) (-3 (Sqrt[2]) Cos[θ]+(Sqrt[6]) (Sqrt[16+3 ((Cos[θ])^2)]))))(∫_(ρ^2)^((1/6) (4-3 (Sqrt[2]) ρ Cos[θ])))(1/(Sqrt[8]))(1/(Sqrt[7]))ρ\[DifferentialD]z\[DifferentialD]ρ\[DifferentialD]θ

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