Answer to Question #238276 in Calculus for Anuj

Question

(Section 13.3 and Chapter 14) Let D be the region in R 3 p that lies inside the cone z = x 2 + y 2 above the plane z = 1 and below the hemisphere z = p 4 − x 2 − y 2 . (a) Sketch the region D in R 3 .(b) Express the volume of D as a sum of triple integrals, using cylindrical coordinates.

Answer 1

Answer:

$V=V_1+V_2=\int_0^{2\pi}\int_0^1\int_1^{\sqrt{4-x^2-y^2}}rdzdrd\phi+ \int_0^{2\pi}\int_1^{\sqrt{2}}\int_r^{\sqrt{4-x^2-y^2}}rdzdrd\phi$

Explanation on bounds of integrals

1) Intersection of cone $z=\sqrt{x^2+y^2}$ and plane z=1 is $1=\sqrt{x^2+y^2}$

or $x^2+y^2=1$ This is the circle with radius 1 and it is upper bound by r in V1 and lower bound with respect to r in V2.

2) Intersection of the cone $z^2=x^2+y^2$ and sphere $x^2+y^2+z^2=4$ is line $x^2+y^2=2,\space z=\sqrt{2}$ . This is circle on height $\sqrt{2}$ with radius $r=\sqrt{2}$ and $\sqrt{2}$ is upper bond by r in V2.

3) For V2 z changes from plane z=1 to sphere $z=\sqrt{4-x^2+y^2}$

4) For V1 z changes from cone $z=\sqrt{x^2+y^2}=r$ to upper bound in spere $z=\sqrt{4-x^2-y^2}$ .

5) All regions possess circular symmetry that is why $\phi$ changes from 0 to $2\cdot \pi$