Setting up Double Intergrals to Calculate Volume

[GogumaDork]

Homework Statement

Set up the double integral in rectangular coordinates for calculating the volume.
The solid is created by the equation

Homework Equations

Fubini's Thereom ∫_R∫f(x,y)dA = ∫_a^b∫f(x,y)dydx

The Attempt at a Solution

The shape created by Z = 30 - x² - y² is 3d parabola facing downards.
Since it is bounded by Z = 5, the base shape is a circle at origin with radius 5 since
5 = 30 - x² - y² is equal to x² + y² = 25
As a result x is bounded by -5 and 5 while y is bounded by -√(25-x^2) and √(25-x^2)

As a result to set up the double integral
∫_-5⁵∫_√(25-x^2)^{√(25-x^2)}30-x^{^2}-y^{^2}dydx

Are the integral boundaries in correct?

 

[uart]

Homework Statement

Set up the double integral in rectangular coordinates for calculating the volume.
The solid is created by the equation

Homework Equations

Fubini's Thereom ∫_R∫f(x,y)dA = ∫_a^b∫f(x,y)dydx

The Attempt at a Solution

The shape created by Z = 30 - x² - y² is 3d parabola facing downards.
Since it is bounded by Z = 5, the base shape is a circle at origin with radius 5 since
5 = 30 - x² - y² is equal to x² + y² = 25
As a result x is bounded by -5 and 5 while y is bounded by -√(25-x^2) and √(25-x^2)

As a result to set up the double integral
∫_-5⁵∫_-√(25-x^2)^{√(25-x^2)}30-x^{^2}-y^{^2}dydx

Are the integral boundaries in correct?

Yes the boundaries are correct (except for the typo where you left out the "-" in the lower limit).

However, you really should be making use of the cylindrical symmetry here and changing to cylindrical coordinates. It makes the problem much easier.

 

[GogumaDork]

Thanks for the response! However, that is the answer I entered into Webwork but it states it is incorrect.

 

[GogumaDork]

Okay I managed to figure it out after reading: http://mathinsight.org/double_integral_volume

My boundaries were correct but the equation I was integrating was not.
It explained that the integral equation is not f(x,y) but f(x,y) - g(x,y), in my case g(x,y) being the "other surface" or z = 5

As a result I subtracted 5 from f(x,y) which was 30 - x^2 - y^2 and resulted in a final answer of
∫_-5⁵∫_-√(25-x^2)^{√(25-x^2)} 25-x^2-y^2dydx

 

[HallsofIvy]

Yes, what you had was the volume between the surface z= 30- x^2- y^2 and the xy-plane, z= 0. Unfortunately, so many basic problem ask for just the "area below the curve" and "the volume below the surface", we may develop the very bad habit of automatically assuming a lower bound of 0.

By the way, in cylindrical coordinates, that uart referred to, the upper surface is given by z= 30- r^2[/tex], the lower surface by z= 5, and the bounding circe by r= 5. The integral would be<br /> \int_{\theta= 0}^{2\pi}\int_{r= 0}^5 (30- r^2- 5) rdrd\theta=\int_{\theta= 0}^{2\pi}\int_{r= 0}^5 (25- r^2) rdrd\theta