Sketch the solid whose volume is given by the specified integral. ∫0^1 ∫0^2(9-x^2-y^2) d y d x

Name: Problem 27, Chapter 14: University Calculus: Early Transcendentals
Uploaded: 2020-04-24T21:41:59-08:00
Duration: 4 min 22 s
Channel: Bobby Barnes
Description: Sketch the solid whose volume is given by the specified integral. ∫0^1 ∫0^2(9-x^2-y^2) d y d x

Sketch the solid whose volume is given by the specified...

Key Concepts

Sketching the Solid

Sketching the solid involves drawing both the base region and the curve represented by the function that defines the height. The base region, defined by the integration limits, forms the bottom of the solid, while the function provides the elevation of the upper boundary of the solid. Visualizing these together helps in understanding the geometry of the three-dimensional figure.

Double Integral

A double integral extends the concept of integration to functions of two variables over a two-dimensional region. It allows for the computation of quantities like area, volume, and mass by summing up contributions from infinitely small elements, where each element's measure is given by the integrand function.

Region of Integration

The region of integration is the area in the xy-plane over which the integral is computed. In problems like this one, the limits of integration define a rectangle or other shape in the plane, serving as the base or domain over which the function is integrated.

Solid under a Surface

When a function of two variables represents the height of a surface above the xy-plane, the volume of the solid bounded above by this surface and below by the base region is computed using a double integral. The integrand (in this case, 9 - x^2 - y^2) represents the distance from the xy-plane to the surface at each point in the region.

Transcript

00:01 We want to sketch the solid whose volume is given by this integral over here.

00:05 Now, they don't tell us to actually integrate it.

00:07 They just want us to sketch the graph of it.

00:09 So now, normally, this inside is going to be our function f of x, y, which is going to tell us our z component.

00:22 So we're given the function z is equal to 9 minus x squared, minus y squared.

00:28 And this is going to be a paraboloid.

00:38 So we know the shape of it now, and what we can go ahead and do is describe our region.

00:50 So first, y is this first integral.

00:54 So we know that y is going to be bounded between zero and two.

01:00 And then x is this outer integral here so that means x is going to be bounded between 0 and 1 now the volume that we're going to have is everything below this curve here so let's go ahead and sketch this curve first and then we can just shade everything in so let's go ahead and plot at least our endpoints so we can look at f of 0 .0.

01:40 So that's just going to give us 9.

01:42 So this is the point 0 .09.

01:48 And then we can look at f of 02.

01:56 So that's going to give us 5.

02:03 So this is going to be 0 to 5.

02:08 Then we can look at f of 1 ,0, which is going to give us 8.

02:14 So this is going to be 1 .08.

02:18 And then we'd look at f of 1, 2, and that's going to give us 4.

02:32 So since all these are positive, let's just go ahead and sketch our rectangular region first.

02:36 So this is going to be about 1, then this would be 2 units...