Double Integral: Finding Area of Paraboloid Beneath z=2

In summary, the conversation discusses finding the area of the paraboloid z=x²+y² beneath z=2. A polar change of variable is used to simplify the integral, and the change of variable \xi=1+4r^2 is used to find the area. The correct answer is found to be 13/3, which is also confirmed by a different method used by the speaker's friends. A mistake is spotted in the original post, but the correct answer is still found to be 13/3.
  • #1
quasar987
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It's about finding the area of the paraboloid z=x²+y² beneath z=2. The area integral is

[tex]\int\int_{\{(u,v): u^2+v^2<2\}}\sqrt{1+4(u^2+v^2)}dudv[/tex]

A polar change of variable seems to fits nicely:

[tex]=\int_0^{\sqrt{2}}\int_0^{2\pi}\sqrt{1+4r^2}rd\theta dr[/tex]

Then the change of variable [itex]\xi=1+4r^2[/itex] brings the area to

[tex]2\pi\int_1^9\sqrt{\xi}\frac{d\xi}{8}=\frac{2\cdot 2 \pi}{3 \cdot 8}[\xi^{3/2}]_1^9=\frac{13\pi}{3}[/tex]

Looks good?
 
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  • #2
You got the right answer. However, you lost a factor of two on the last line of math and somehow gained that factor of two back again to get the correct answer. Those equalities are not equalities as written.
 
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  • #3
My friends found the area by "setting" z=u²+v² and just evaluating

[tex]=\int_0^{2}\sqrt{1+4z}dz[/tex]

They found 13/3 (with no pi). I really can'T find any kind of justification for this integral but they say this is how the teaching assistant did a similar problem.

Any ways, thanks for confirming D H.

Btw - I think I just spotted a mistake in my OP. Instead of

[tex]2\pi\int_1^9\sqrt{\xi}\frac{d\xi}{8}=\frac{2\pi}{3 \cdot 8}[\xi^{3/2}]_1^9=\frac{13\pi}{3}[/tex]

it should be

[tex]2\pi\int_1^9\sqrt{\xi}\frac{d\xi}{8}=\frac{2\cdot 2 \pi}{3 \cdot 8}[\xi^{3/2}]_1^9=\frac{26\pi}{3}[/tex]

(I edited the OP)
 
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  • #4
D H said:
You got the right answer. However, you lost a factor of two on the last line of math and somehow gained that factor of two back again to get the correct answer. Those equalities are not equalities as written.

Oh I see. Ok, we're back as 13/3, but with the right amount of two's in the last line.
 

1. What is a double integral?

A double integral is a type of integral in multivariable calculus that involves integrating over a two-dimensional region. It is used to calculate the volume, surface area, and other quantities of three-dimensional objects.

2. How do you find the area of a paraboloid beneath z=2 using a double integral?

To find the area of a paraboloid beneath z=2 using a double integral, you first need to set up the integral in terms of x and y. This can be done by expressing the equation of the paraboloid as z=f(x,y) and then integrating the function over the region of interest. The result will be the area of the paraboloid beneath z=2.

3. What is the region of integration for a double integral of a paraboloid beneath z=2?

The region of integration for a double integral of a paraboloid beneath z=2 is typically a circular or elliptical region in the xy-plane. The boundaries of the region can be determined by setting the equation of the paraboloid equal to 2 and solving for either x or y.

4. Can a double integral be used to find the volume of a paraboloid?

Yes, a double integral can be used to find the volume of a paraboloid. By setting the limits of integration appropriately, the double integral can be used to calculate the volume of the region under the paraboloid. This is because the volume can be thought of as the sum of the areas of infinitely thin slices of the paraboloid, which can be calculated using the double integral.

5. Are there any applications of double integrals in real life?

Yes, double integrals have many applications in real life, particularly in physics and engineering. They are used to calculate the mass, center of mass, and moments of inertia of three-dimensional objects. They are also used in modeling fluid flow and heat transfer in various systems. Additionally, double integrals are used in statistics to calculate probabilities and expected values of continuous variables.

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