Quick double integral question

[STEMucator]

Homework Statement

Find the volume of the region R between the surfaces $z = 4x^2 + 2y^2 \space and \space z = 3 + x^2 - y^2$

Homework Equations

The Attempt at a Solution

Okay so I think I have an idea about how to do this one. First I check when the two surfaces intersect, that is when $4x^2 + 2y^2 = 3 + x^2 - y^2$. So the two surfaces meet when $x^2 + y^2 = 1$ which is a circle of radius 1 with center (0,0).

So either x or y run from -1 to 1, it doesn't really matter which one I pick here I think, so I'll hold y fixed : $-1 ≤ y ≤ 1$

Now solving $x^2 + y^2 = 1$ for x yields $x = ± \sqrt{1-y^2}$ as my upper and lower limits for x.

Now all that's left to figure out is... what is f? I believe intuitively that f will be the bigger surface minus the smaller surface. So subtracting my two surfaces I get : $3 + x^2 - y^2 - 4x^2 - 2y^2 = 3(1 - x^2 -y^2)$

So that my iterated integral becomes :

$\int_{-1}^{1} \int_{- \sqrt{1-y^2}}^{\sqrt{1-y^2}} 3(1 - x^2 -y^2) \space dxdy$

Evaluating that shouldn't be a problem, I'm just hoping I did everything properly while setting it up. If anyone could clarify for me it would be great :)!

 

[Mark44]

Homework Statement

Find the volume of the region R between the surfaces $z = 4x^2 + 2y^2 \space and \space z = 3 + x^2 - y^2$

Homework Equations

The Attempt at a Solution

Okay so I think I have an idea about how to do this one. First I check when the two surfaces intersect, that is when $4x^2 + 2y^2 = 3 + x^2 - y^2$. So the two surfaces meet when $x^2 + y^2 = 1$ which is a circle of radius 1 with center (0,0).

So either x or y run from -1 to 1, it doesn't really matter which one I pick here I think, so I'll hold y fixed : $-1 ≤ y ≤ 1$

Now solving $x^2 + y^2 = 1$ for x yields $x = ± \sqrt{1-y^2}$ as my upper and lower limits for x.

Now all that's left to figure out is... what is f? I believe intuitively that f will be the bigger surface minus the smaller surface. So subtracting my two surfaces I get : $3 + x^2 - y^2 - 4x^2 - 2y^2 = 3(1 - x^2 -y^2)$

This looks OK. The surface z = 4x² + 2y² has a larger z value than the surface z = 3 + x² - y² for most points in the domain; namely all of the points that are outside the circle of intersection. The region you're interested in, though, is on or inside the circle. For this region, the second equation has the larger z values.

So that my iterated integral becomes :

$\int_{-1}^{1} \int_{- \sqrt{1-y^2}}^{\sqrt{1-y^2}} 3(1 - x^2 -y^2) \space dxdy$

Evaluating that shouldn't be a problem, I'm just hoping I did everything properly while setting it up. If anyone could clarify for me it would be great :)!

I think it would be much easier to convert to polar form. The circle simplifies to r = 1, and the integrand simplifies to 3(1 - r²).

 

[mfb]

f(x,y)=1-x²-y² gives a [strike]sphere[/strike] paraboloid, you can use this to avoid integrals at all (and add the prefactor later). Otherwise, polar coordinates are useful.

Edit: Oh, missed the missing square root.

 

[STEMucator]

I don't believe we're allowed to use polar coordinates yet ( wish I could ), but I would see why it would make things easier.

So my integrand is correct, but incorrect? A little confused with the feedback.

 

[Mark44]

f(x,y)=1-x²-y² gives a sphere

No, it doesn't. The graph of this function is a cone. If the equation were z² = 1-x²-y², then the graph would be a sphere.

, you can use this to avoid integrals at all (and add the prefactor later). Otherwise, polar coordinates are useful.

 

[Mark44]

I don't believe we're allowed to use polar coordinates yet ( wish I could ), but I would see why it would make things easier.

So my integrand is correct, but incorrect? A little confused with the feedback.

I believe your integrand is correct. I don't believe anyone said it was incorrect.