Find the volumes of bodies bounded by the surfaces: 1) z=x^2+4 y^2, z=2; 2) z=x^2+y^2 / 4-z^2=1,

Find the volumes of bodies bounded by the surfaces: 1)...

Key Concepts

Integrating over Regions Bounded by Quadratic Surfaces

Many volume problems involve regions bounded by quadratic surfaces, such as paraboloids, ellipsoids, and hyperboloids. Recognizing these surfaces allows one to use appropriate coordinate systems and integration strategies to handle the curved boundaries, often by scaling variables or utilizing symmetry in the geometric shapes.

Change of Coordinates

Changing coordinates, such as switching to cylindrical, spherical, or elliptical coordinate systems, can simplify the integration process when the region exhibits symmetry or has boundaries given by quadratic equations. This technique also involves the use of Jacobian determinants to correctly transform the volume element from one coordinate system to another.

Triple Integrals

Triple integrals are used to calculate the volume of a three-dimensional region. They involve integrating a function (often 1 when computing volume) over a region in space, with integration limits defined by the bounding surfaces. Setting up these integrals requires a careful understanding of the region’s boundaries in all three dimensions.

Determining Limits of Integration

Determining the limits of integration involves analyzing the intersections of the bounding surfaces to accurately describe the region of interest. This process requires solving the equations defining the surfaces, and expressing one variable in terms of others if necessary, to set proper integration bounds that cover the entire volume.

Transcript

00:02 All right, z equals x squared plus y squared plus 3 is up above the x, y, y, plane, also z equals 1 is above there.

00:11 So those are not going to affect the area that we're going to integrate over.

00:15 So the volume is just going to be z on the top, which is x squared plus y squared plus 3, minus z on the bottom, which is one over this area of this region.

00:28 And the region looks like y equals x squared, which is a parabola.

00:33 To y equals four okay so we're gonna i'm gonna integrate this way so z z equals x squared to z equals four from x equals negative 2 to 2 because this is negative 2 to 2 x squared plus y squared plus 2 d y d x so negative 2 x squared to 2 2x2 from x squared to 4 d x negative 2 to 2 2 4 d x negative 2 to 2 4x squared plus 4 cubed which is 64 3rds plus 8 minus x to the 4th plus x to the 6th over 3 plus 2 x squared d x let's go ahead and combine some like terms there before we integrate.

01:51 Here we have 4x squared minus 2x squared, so 2x squared.

01:57 64 thirds plus 24 thirds, that'd be 88 thirds, minus x to the fourth, minus x to the 6th over 3, dx.

02:14 So 2x cubed over 3 plus 88 thirds x, minus x to the 5 over 5, minus x to the 7 over 21 from negative 2 to 2.

02:31 So 16 thirds plus 176 thirds minus 32 5ths plus 128 over 21 minus.

02:47 So the exact same thing, only opposites, because these are all odd functions.

03:05 Minus plus plus.

03:08 Okay, that's right.

03:10 So 32 thirds, oops, 176 there, plus 352 thirds, minus 64 fifths...