The volume cut from the solid bounded $x ^ { 2 } + 4 y ^ { 2 } = 4$ by z = 0 and z = x + 2 is

A) $4 \pi$
B) $3 \pi$
C) $2 \pi$
D) $\frac { 4 \pi } { 3 }$
E) $\frac { 3 \pi } { 2 }$

Question

The volume cut from the solid bounded $x ^ { 2 } + 4 y ^ { 2 } = 4$ by z = 0 and z = x + 2 is

A) $4 \pi$
B) $3 \pi$
C) $2 \pi$
D) $\frac { 4 \pi } { 3 }$
E) $\frac { 3 \pi } { 2 }$

Quizplus · Accepted Answer

The given solid is an elliptical cylinder with the base ellipse $x^2 + 4y^2 = 4$ and height given by the plane $z = x + 2$. To find the volume cut from this solid, we can integrate the area of cross-sections perpendicular to the z-axis. The base ellipse can be rewritten as $\frac{x^2}{2^2} + \frac{y^2}{1^2} = 1$, indicating it's an ellipse with semi-major axis 2 and semi-minor axis 1. The volume of the solid can be found by integrating the area of the ellipse along the z-axis from z=0 to z=2 (the maximum value of z=x+2 when x=0). The area of the ellipse is $A = \pi ab = \pi(2)(1) = 2\pi$. Since the height of the solid (from z=0 to z=2) does not depend on x or y, the volume is simply the area of the base times the height, which is $2\pi \times 2 = 4\pi$.

The Volume Cut from the Solid Bounded x2+4y2=4x ^ { 2 } + 4 y ^ { 2 } = 4x2+4y2=4

The Volume Cut from the Solid Bounded $x ^ { 2 } + 4 y ^ { 2 } = 4$