21-32 Use spherical coordinates. Evaluate ∫𝕁H(9-x^2-y^2) d V, where H is the solid hemisphere x^

21-32 Use spherical coordinates. Evaluate ∫𝕁H(9-x^2-y^2) d...

Key Concepts

Region of Integration Transformation

Transforming the region of integration when changing coordinate systems involves rewriting the geometric boundaries using the new variables. For spherical coordinates, this means expressing conditions (like being part of a hemisphere) in terms of ?, ?, and ?, which can simplify the specification of integration limits.

Triple Integration

Triple integration involves computing the integral of a function over a three-dimensional region. When the region exhibits spherical symmetry, converting to spherical coordinates simplifies the integration process by aligning the integration limits with the geometry of the region.

Jacobian Determinant

The Jacobian determinant is crucial in transforming integrals when changing variables. In spherical coordinates, the volume element dV becomes ?² sin(?) d? d? d?, accounting for the change in scale from Cartesian to spherical coordinates.

Spherical Coordinates

Spherical coordinates are a system in which points in three-dimensional space are represented by a distance from the origin and two angles. This coordinate system is particularly useful for regions with spherical symmetry, such as spheres and hemispheres, because the variables often separate the geometry of the region.

Transcript

00:01 Okay, so we've got to use spherical coordinates here, and we can go ahead and just use what we know.

00:09 So we know that in spherical coordinates we want to rewrite this function of x, y.

00:17 So we know that x is just row cosine theta, sine, sine, phi, and y is row sine theta, sine phi.

00:29 So in our case, we know that i won't write the bounds yet.

00:39 So we'll have 9 minus row squared, cosine squared, sine squared, row squared, sine squared, sine squared, and we'll end up with row squared factored out and a sine squared factor out as well.

00:56 Sign squared of five factored out.

00:59 Cosine squared theta plus sine squared theta dv.

01:04 And we know this part is one, so we'll have the triple 1 roll over h of 9 minus row squared sine squared phi dv.

01:16 And now we can go ahead and continue to substitute.

01:21 Turning dv into row phi and theta gives us dv is equal to row squared sine phi, d row, d -rode, d -fi, d -theta.

01:31 And then we've got to think about what our solid is.

01:35 So we have a solid h.

01:36 It's the hemisphere, x squared plus y square plus z squared is less than nine above the z equal zero plane.

01:47 So this means that it's the top half of a hemisphere of radius three.

01:54 So we know that our bounds should be reflective of this.

01:58 So we have, we'll write our plane, so we have x, y, z, or space.

02:07 So we have a hemisphere, and then, so it's this top hemisphere of radius 3.

02:17 So we have the following bounds.

02:22 Row is between 0 and 3.

02:25 Since it's a top hemisphere, we have 5 between 0 and pi over 2, and since it goes all the way around, data is just between 0 and 2 pi.

02:36 So we end up with the integral of 0 to 2 pi.

02:42 0 to pi over 2, 0 to 3, of 9 minus row squared sine squared phi, row squared sine, phi, d, phi, d, phi, d theta.

03:01 Now this becomes a little more complicated, but we'll go ahead and tackle this quickly.

03:09 So we'll get, we'll keep the outer intervals, and then we'll notice that we'll get, 9 row squared sine phi, and then minus 9 row to the fourth sine cubed phi.

03:34 And then this is all included, d row, d, phi, d theta.

03:39 Now, i forgot to write the inner bound, all the bounds, so we'll go ahead and write that as well.

03:53 So then we'll end up with total these three bounds...

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