# Double integral of volume bounded by plane and paraboloid

1. Sep 26, 2007

Evaluate the volume of the solid bounded by the plane z=x and the paraboloid z = x^2 + y^2

I have tried to graph this, and they don't bound anything? have i graphed it wrong. and is there a way to do these problems where you don't need to draw the graph.

2. Sep 27, 2007

### HallsofIvy

Staff Emeritus
They surely do bound something! If the 3-d graph bothers you assume y= 0. Then your graphs are z= x and z= x^2, which intersect at (0,0) and (0,1).
Letting z= x in z= x^2+ y^2, you get x^2- x+ y^2= x^2- x+ 1/4 + y^2= 1/4 so
(x-1/2)^2+ y^2= 1/4. In the xy-plane, that's a circle with center at (1/2, 0) and tangent to the y-axis. That's the projection of the actual intersection in the xy-plane.

Last edited: Sep 27, 2007
3. Sep 27, 2007

### EnumaElish

Graphs can be very useful in this type of problem.

z = x is like a ramp that runs the length of the y axis.

What does the parabola equation describe? Can you visualize it?

4. Sep 27, 2007

### HallsofIvy

Staff Emeritus
This was also posted in the homework forum. I'm merging it into that thread.

5. Sep 27, 2007

am i able to change to polar coordinates with the shifted circle? or should i keep it in x y form

let me know if this is correct:
$$\int^{1/2}_{0}\int^{\sqrt{1/4-(x-1/2)^{2}}}_{-\sqrt{1/4-(x-1/2)^{2}}} x^{2} + y^{2} dy dx$$

I think i want to convert to polar form.. but i don't know what to do with the 1/4 inside the ( ) with the x when converting.
also, is my double integral even correct?

6. Sep 27, 2007

### HallsofIvy

Staff Emeritus
If it were me, I would convert to polar coordinates. If $(x- 1/2)^2+ y^2= 1/4$, then $x^2+ y^2- x+ 1/4= 1/4$ or $x^2+ y^2- x= 0$ so, in polar coordinates $r^2- rcos(\theta)= 0$ and, finally, $r= cos(\theta)$. Looks nice!