## Related questions with answers

Question

Sketch the solid described by the given inequalities. 0<=r<=2, -pi/2<=theta<=pi/2, 0<=z<=1

Solutions

VerifiedSolution A

Solution B

Answered 1 year ago

Step 1

1 of 3First focus on $r\leq z\leq2$

$r = z$ is $z = \sqrt{x^2+y^2}$, this is a cone

$z=2$ is a plane.

$r\leq z\leq 2$ is region below the plane $z=2$, but above the cone $z=r$

Answered 1 month ago

Step 1

1 of 2From inequalities $r \le z \le 2$ We have: $\sqrt{x^2+y^2} \le z \le 2$ Let's sketch this, on the left side of this inequality we have a cone, and on the right side we are looking below the plane.

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