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

Answer:

The volume of the solid is: $\mathbf{\dfrac{76}{3}}$

Step-by-step 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)| 0 ≤ x ≤ y, 0 ≤ y ≤ 2}

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

D = {(x,y) | a ≤ y  ≤ b, g₁(y)  ≤ x  ≤ g₂ (y) }

Then, we can compute the double integral as follows:

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

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

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

$\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_{0} (7\dfrac{x^3}{3}+4xy^2)^y_0 \ dy$

$\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_{0} (7\dfrac{y^3}{3}+4(y)y^2-0) \ dy$

$\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_{0} (7\dfrac{y^3}{3}+4y^3) \ dy$

$\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_{0} ( \dfrac{7y^3+ 12y^3}{3} \ ) \ dy$

$\int \int_D (7x^2 +4y^2) \ dA = \int \limits ^2_{0} ( \dfrac{ 19y^3}{3} \ ) \ dy$

$\int \int_D (7x^2 +4y^2) \ dA = \dfrac{ 19}{3} \int \limits ^2_{0} y^3 \ dy$

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

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

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

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

$\int \int_D (7x^2 +4y^2) \ dA = \dfrac{ 76}{3}$

Thus, the volume of the solid is: $\mathbf{\dfrac{76}{3}}$