Find the volume of the solid bounded below by the parabloid z = x^2 + y^2 and above by the plane 2x + z = 3
This is just
∫∫∫ dz dy dx using the appropriate limits
Unfortunately, the plane 2x+z=3 is a vertical plane, and does not bound a region of the paraboloid.
Ok! Thanks for your help :)
To find the volume of the solid bounded by the paraboloid z = x^2 + y^2 and the plane 2x + z = 3, we need to evaluate the surface integral $\iint_S dS$ over the region where the two surfaces intersect.
Step 1: Set up the region of integration in the xy-plane.
- We want to find the intersection between the paraboloid and the plane, so we set the two equations equal to each other:
x^2 + y^2 = 3 - 2x
Rearranging the equation, we get:
x^2 + 2x + y^2 - 3 = 0
- This equation represents a circle centered at (-1, 0) with a radius of 2. So the region of integration in the xy-plane is a circle of radius 2 centered at (-1, 0).
Step 2: Determine the bounds for integration.
- In polar coordinates, the equation of the circle becomes:
r^2 = 3 - 2r cos(theta) - r^2 sin^2(theta)
2r cos(theta) = 3 - 2r - r^2 sin^2(theta)
r = (3 - 2r - r^2 sin^2(theta))/(2 cos(theta))
- To determine the bounds for r, we find the values of r where the above equation is satisfied for each value of theta within the region of integration (the circle).
- To find the bounds for theta, we consider the angle that sweeps from the positive x-axis to the circle. Since the circle is centered at (-1, 0), we rotate it by π radians to shift its center to the origin, and the angle theta will be measured with respect to the positive x-axis in the range [0, 2π].
Step 3: Evaluate the integral.
The volume of the solid is given by the double integral of 1 over the region of integration in polar coordinates:
V = ∫∫ r dr dθ
- The inner integral with respect to r has bounds determined in Step 2, where r ranges from the innermost point of the region to the outermost point for each value of theta.
- The outer integral integrates over the range of theta from 0 to 2π.
Step 4: Calculate the volume.
Evaluate the double integral using the determined bounds and integrate:
V = ∫[0,2π] ∫[r1, r2] r dr dθ
Once you evaluate this integral, you will obtain the volume of the solid bounded by the given surfaces.