# [MultiVarCalc] Find the volume of the solid

• bornofflame
In summary, the volume of the solid between the cone and paraboloid is given by the integral of r from 0 to 3, z from r to 12 - r^2, and theta from 0 to 2pi, which evaluates to 99/2*pi.
bornofflame

## Homework Statement

Find the volume of the solid between the cone ##z = \sqrt{x^2 + y^2}## and the paraboloid ##z = 12 - x^2 - y^2##.

## Homework Equations

##x^2 + y^2 = r^2##

## The Attempt at a Solution

I drew a simple diagram to start off with to visualize the solid formed by the intersection of these equations which is in the attached image along with my work in raw form.

Since we're dealing with circular cross sections the polar coordinates will be best, (read easiest), to work with for me.

Volume = ?
Setting up the integral
##V = \int_R dV = \int_R r\, dz\, dr\, d\theta##
## z = \sqrt{x^2 + y^2} = \sqrt{r^2} = r##
## z = 12 - x^2 - y^2 = 12 - r^2##

lower limit for z ## = r##
upper limit for z ## = 12 - r^2##

##\sqrt{x^2 + y^2} = 12 - x^2 - y^2##
##r = 12 - r^2##
##r^2 + r - 12 = 0##
##(r + 5)(r - 3) = 0##
##r = -4, 3: -4 \text{ is out of bounds, so }r = 3##
## 0 \leq r \leq 3##

##V = \int_0^{2\pi} \int_0^3 \int_r^{r^2} r\, dz\, dr\, d\theta##
##= \int_0^{2\pi} \int_0^3 rz\, \left. \right|_r^{r^2}\, dr\, d\theta##
##= \int_0^{2\pi} \int_0^3 r^3-r^2\,dr\,d\theta##
##=\int_0^{2\pi} \frac 1 4 r^4- \frac 1 3 r^3 \left. \right|_0^3 \,d\theta##
##=\int_0^{2\pi} \frac 81 4 - 9\, d/theta = \frac 81 4 \theta - 9\theta \left. \right|_0^{2\pi}##
##=\frac {81} 4 (2\pi)-9(2\pi) = \frac {81} 2\pi - 18\pi = \frac {81} 2\pi - \frac {36} 2 \pi##
##=\frac {45} 2 \pi##

#### Attachments

• Math263_review.jpg
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Last edited:
It looks like the upper limit in the innermost integral, the one over ##z##, is wrong. It needs to be the ##z## vertical coordinate of the paraboloid where the horizontal coords are ##(r,\theta)##. That will not be ##r^2##.

bornofflame
Should the upper limit be ##12 - r^2##?

Which would give me ##\frac {99} 2 \pi## as an answer instead.

Last edited:
bornofflame said:
Should the upper limit be ##12 - r^2##?
Yes

## What is the formula for finding the volume of a solid?

The formula for finding the volume of a solid is V = ∫A(x)dx, where A(x) is the cross-sectional area of the solid at a given point x.

## How do you determine the limits of integration for finding the volume of a solid?

The limits of integration for finding the volume of a solid are determined by the range of values for x that make up the solid. These limits can be determined by examining the given solid or by using the limits of the given function.

## Can the volume of a solid be negative?

No, the volume of a solid cannot be negative. Volume is a measure of the amount of space occupied by an object, and it is always a positive value.

## What is the difference between finding the volume of a solid using single variable calculus and multivariable calculus?

In single variable calculus, the volume of a solid can be found by integrating the area function over a single variable. In multivariable calculus, the volume is found by integrating the cross-sectional area function over multiple variables, which allows for more complex shapes to be calculated.

## Are there any special cases when finding the volume of a solid?

Yes, there are special cases when finding the volume of a solid, such as when the solid has holes or when the cross-sectional area changes at a certain point. In these cases, the volume can be found by breaking the solid into smaller, simpler shapes and integrating over each individual shape.

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