Can someone check my integral real quick ?

  • #1

Homework Statement


use a double integral to find the volume bounded by the paraboloid :[tex]z=4-x^2-y^2[/tex], xy-plane and inside a cylinder: [tex]x^2+y^2=1[/tex]

Homework Equations



x=rcosθ y=rsinθ

The Attempt at a Solution


the radius of the area of integration is 1, since its determined by the cylinder only, and the cylinder has radius of 1.
the cylinder has an infinite z value, so Z is like like [tex]4-r^2- 0[/tex]
so I got this:

[tex] \int_{θ=0}^{2π} \int_{r=0}^1 (4-r^2)r \, dr \, dθ[/tex]
 

Answers and Replies

  • #2
34,159
5,778

Homework Statement


use a double integral to find the volume bounded by the paraboloid :[tex]z=4-x^2-y^2[/tex], xy-plane and inside a cylinder: [tex]x^2+y^2=1[/tex]

Homework Equations



x=rcosθ y=rsinθ

The Attempt at a Solution


the radius of the area of integration is 1, since its determined by the cylinder only, and the cylinder has radius of 1.
the cylinder has an infinite z value, so Z is like like [tex]4-r^2- 0[/tex]
so I got this:

[tex] \int_{θ=0}^{2π} \int_{r=0}^1 (4-r^2)r \, dr \, dθ[/tex]
That's it.

When you actually get to where you're evaluating these integrals, rather than just setting them up, there are some shortcuts you can take. Due to the symmetry of the two bounding surfaces, you can find the volume in the first octant, and multiply that by 4 to get the entire volume. IOW, this:
##4 \int_{θ=0}^{π/2} \int_{r=0}^1 (4-r^2)r \, dr \, dθ##
 
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  • #3
That's it.

When you actually get to where you're evaluating these integrals, rather than just setting them up, there are some shortcuts you can take. Due to the symmetry of the two bounding surfaces, you can find the volume in the first octant, and multiply that by 4 to get the entire volume. IOW, this:
##4 \int_{θ=0}^{π/2} \int_{r=0}^1 (4-r^2)r \, dr \, dθ##
gotcha, thank you
 

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