Question

Find the volume of the given solid.Enclosed by the paraboloidz = 7x2 + 4y2and the planesx = 0, y = 2, y = x, z = 0

Answer 1

The volume of the solid is:
`\mathbf{(76)/(3)}`

Explanation:

Consider the given paraboloid and the planes:

z = 7x² + 4y², x = 0, y = 2, y = x , z = 0

The region of type -II can be expressed as:

D = (x,y)

Suppose f(x,y) is continous on type-I region D such that:

D = a ≤ y ≤ b, g₁(y) ≤ x ≤ g₂ (y)

Then, we can compute the double integral as follows:

`\int \int \limits _D f (x,y) \ dA = \int \limits ^b_a \int \limits ^(g_2(x))_(g_1 (y)) f (x,y) \ dydx`

`\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_(y=0) \int \limits ^(y)_(x=0)(7x^2+4y^2) \ dxdy`

`\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_(0) (7x^2+4y^2)^y_0 \ dy`

`\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_(0) (7(x^3)/(3)+4xy^2)^y_0 \ dy`

`\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_(0) (7(y^3)/(3)+4(y)y^2-0) \ dy`

`\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_(0) (7(y^3)/(3)+4y^3) \ dy`

`\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_(0) ( (7y^3+ 12y^3)/(3) \ ) \ dy`

`\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_(0) ( ( 19y^3)/(3) \ ) \ dy`

`\int \int_D (7x^2 +4y^2) \ dA = ( 19)/(3) \int \limits ^2_(0) y^3 \ dy`

`\int \int_D (7x^2 +4y^2) \ dA = ( 19)/(3) \begin {bmatrix} (y^4)/(4) \end {bmatrix}^2_0`

`\int \int_D (7x^2 +4y^2) \ dA = ( 19)/(3) \begin {bmatrix} (2^4)/(4) -0 \end {bmatrix}`

`\int \int_D (7x^2 +4y^2) \ dA = ( 19)/(3) \begin {bmatrix} (16)/(4) -0 \end {bmatrix}`

`\int \int_D (7x^2 +4y^2) \ dA = ( 19)/(3) \begin {bmatrix} 4\end {bmatrix}`

`\int \int_D (7x^2 +4y^2) \ dA = ( 76)/(3)`

Thus, the volume of the solid is:
`\mathbf{(76)/(3)}`