Find the volume of the solid that is below the cone z= sqrt(x2 + y2). and above the region 1 <or= x2+y2 <or= 4 Find the volume of the solid that is below the cone z = Vx2 + y2 and above the region 1<x²+y² < 4.

Answered: Find the volume of the solid that is…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Find the volume of the solid that is below the cone z= sqrt(x²+ y²). and above the region 1 <or= x²+y² <or= 4

Find the volume of the solid that is below the cone z = Vx2 + y2 and above the region 1<
x²+y² < 4.

Expert Solution

Step 1

The domain D is a polar rectangle bounded by concentric circles r = 1 and r = 2 with 0 ≤ θ ≤ 2π.

We know that, $x = r \cos θ$ , $y = r \sin θ$ , and $d A = (r) d r d θ$ in polar form. The equation of the cone becomes $z = r$ . So the integral becomes as

$\begin{array}{rcl} \iint_{D} \sqrt{x^{2} + y^{2}} d A & = & \int_{0}^{2 π} \int_{1}^{2} r (r d r) d θ \\ = & \int_{0}^{2 π} \int_{1}^{2} r^{2} d r d θ \\ = & \int_{0}^{2 π} {(\frac{r^{3}}{3})}_{1}^{2} d θ \\ = & \int_{0}^{2 π} (\frac{{(2)}^{3}}{3} - \frac{{(1)}^{3}}{3}) d θ \\ = & \frac{7}{3} \int_{0}^{2 π} d θ \\ = & \frac{7}{3} (2 π - 0) \\ = & \frac{14 π}{3} \end{array}$

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