Find the volume under the paraboloid $z = 4 x ^ { 2 } + y ^ { 2 }$ above the triangle with vertices (0, 0, 0), (3, 0, 0), and (3, 1, 0).
A) $\frac { 109 } { 4 }$

B) $\frac { 55 } { 2 }$
C) $\frac { 111 } { 4 }$
D)28
E) $\frac { 113 } { 4 }$
F) $\frac { 57 } { 2 }$

G) $\frac { 115 } { 4 }$
H)29

Question

Find the volume under the paraboloid $z = 4 x ^ { 2 } + y ^ { 2 }$ above the triangle with vertices (0, 0, 0), (3, 0, 0), and (3, 1, 0).
A) $\frac { 109 } { 4 }$ 
 
B) $\frac { 55 } { 2 }$ 
C) $\frac { 111 } { 4 }$
D)28
E) $\frac { 113 } { 4 }$
F) $\frac { 57 } { 2 }$ 
 
G) $\frac { 115 } { 4 }$ 
H)29

Quizplus · Accepted Answer

To find the volume under the paraboloid $z = 4x^2 + y^2$ above the given triangle, we use a double integral over the triangular region in the xy-plane. The triangle's vertices suggest it spans from $x = 0$ to $x = 3$ and $y = 0$ to $y = 1$, with a linear relation between $y$ and $x$ for the hypotenuse. The volume can be calculated as $\int_0^3 \int_0^x (4x^2 + y^2) dy dx$. Solving this integral gives the volume under the paraboloid over the triangular region, which is $\frac{109}{4}$.

Find the Volume Under the Paraboloid z=4x2+y2z = 4 x ^ { 2 } + y ^ { 2 }z=4x2+y2

Find the Volume Under the Paraboloid $z = 4 x ^ { 2 } + y ^ { 2 }$