Finding the volume between two surfaces

[supermiedos]

Homework Statement

Find the volume between z = x^2 + y^2 and z = 3 - x - y

Homework Equations

None

The Attempt at a Solution

I must use a double integral. Using polar coordinates I find that the volume is equal to:

V = ∫∫(3 - rcosθ - rsinθ) r dr dθ - ∫∫r^2 r dr dθ

I'm struggling trying to find the region of integration.
I found that the projection onto the x-y plane is the circle x^2 + x + y^2 + y = 3, or (x+1/2)^2+(y+1/2)^2 = 7/2. By switching to polar coordinates I get: r^2 + rcos + rsinθ = 3

Since it's a circle, I assume that 0 ≤ θ ≤ 2∏.
But what about r?

I tought It went from 0 <= r <= √(7/2), but I'm not sure anymore since it's a circle centered outside the origin. What can I do?

 

[LCKurtz]

Homework Statement

Find the volume between z = x^2 + y^2 and z = 3 - x - y

Homework Equations

None

The Attempt at a Solution

I must use a double integral. Using polar coordinates I find that the volume is equal to:

V = ∫∫(3 - rcosθ - rsinθ) r dr dθ - ∫∫r^2 r dr dθ

I'm struggling trying to find the region of integration.
I found that the projection onto the x-y plane is the circle x^2 + x + y^2 + y = 3, or (x+1/2)^2+(y+1/2)^2 = 7/2. By switching to polar coordinates I get: r^2 + rcos + rsinθ = 3

Since it's a circle, I assume that 0 ≤ θ ≤ 2∏.
But what about r?

I tought It went from 0 <= r <= √(7/2), but I'm not sure anymore since it's a circle centered outside the origin. What can I do?

Those limits would be correct if the circle was centered at the origin, and $r$ and $\theta$ were measured from there. Your integral is $$
\iint_R 3-x-y-(x^2+y^2)dydx =\iint_R 3-(x+y+x^2+y^2)dydx=\iint_R\frac 7 2 - (x+\frac 1 2)^2-(y+\frac 1 2)^2 dydx$$Now if you let $x+\frac 1 2 = r\cos\theta$ and $y+\frac 1 2 = r\sin\theta$ and $dydx=rdrd\theta$ your limits would be correct. This centers your polar coordinates at the center of the circle.

 

[supermiedos]

Those limits would be correct if the circle was centered at the origin, and $r$ and $\theta$ were measured from there. Your integral is $$
\iint_R 3-x-y-(x^2+y^2)dydx =\iint_R 3-(x+y+x^2+y^2)dydx=\iint_R\frac 7 2 - (x+\frac 1 2)^2-(y+\frac 1 2)^2 dydx$$Now if you let $x+\frac 1 2 = r\cos\theta$ and $y+\frac 1 2 = r\sin\theta$ and $dydx=rdrd\theta$ your limits would be correct. This centers your polar coordinates at the center of the circle.

Let me see if I understand. What you did was translating both surfaces and the projection to (-1/2, -1/2)?

 

[HallsofIvy]

Rephrase that please. It doesn't make sense to talk about moving an entire surface to a single point. Do you mean translate the surfaces so that (-1/2, -1/2) is translated to (0, 0)?

 

[supermiedos]

Rephrase that please. It doesn't make sense to talk about moving an entire surface to a single point. Do you mean translate the surfaces so that (-1/2, -1/2) is translated to (0, 0)?

Yes, sorry. That was what I wanted to say. To make (-1/2, -1/2) the new origin. That's what happened?

 

[LCKurtz]

Yes, sorry. That was what I wanted to say. To make (-1/2, -1/2) the new origin. That's what happened?

I guess you could think of it that way. I prefer to think of it as putting the polar coordinates at the center of the circle. Let $(x,y)$ be any point inside the circle. The position vector $\vec R=\langle x,y\rangle$ goes from the origin to that point. Let $\vec R_0=\langle -\frac 1 2,-\frac 1 2\rangle$ be the vector from the origin to the center of the circle. Now if you let $r,\theta$ be the polar coordinates from the center of the circle to $(x,y)$ then the vector from the center of the circle to $(x,y)$ is $\vec R_1=\langle r\cos\theta,r\sin\theta\rangle$. Now you have $\vec R =\vec R_0+\vec R_1$ or $\langle x,y\rangle=\langle -\frac 1 2,-\frac 1 2\rangle+\langle r\cos\theta,r\sin\theta\rangle$.

 

[supermiedos]

I guess you could think of it that way. I prefer to think of it as putting the polar coordinates at the center of the circle. Let $(x,y)$ be any point inside the circle. The position vector $\vec R=\langle x,y\rangle$ goes from the origin to that point. Let $\vec R_0=\langle -\frac 1 2,-\frac 1 2\rangle$ be the vector from the origin to the center of the circle. Now if you let $r,\theta$ be the polar coordinates from the center of the circle to $(x,y)$ then the vector from the center of the circle to $(x,y)$ is $\vec R_1=\langle r\cos\theta,r\sin\theta\rangle$. Now you have $\vec R =\vec R_0+\vec R_1$ or $\langle x,y\rangle=\langle -\frac 1 2,-\frac 1 2\rangle+\langle r\cos\theta,r\sin\theta\rangle$.

Now I get it. Thank you so much :)