Evaluate ∭E x e^x^2+y^2+z^2 d V, where E is the portion of...

Key Concepts

Spherical Coordinate System

This concept involves converting Cartesian coordinates to spherical coordinates. It is especially useful when dealing with problems where the region of integration is spherical in nature, such as a ball or sphere. By using spherical coordinates, integrals over spherical regions become more manageable due to the natural adaptation of the coordinate system to the geometry of the region.

Jacobian Determinant

When performing a coordinate transformation, the Jacobian determinant accounts for the change in volume element. In spherical coordinates, the volume element is given by the product r²sin(?) dr d? d?, which replaces the Cartesian dV. The Jacobian ensures that the transformation preserves the integral's value by correcting for the stretching and compression of space.

Region of Integration in the First Octant

This concept relates to understanding and defining the limits of integration when the domain is restricted to the first octant. The first octant in three dimensions comprises the region where all the Cartesian coordinates are non-negative. In spherical coordinates, this restriction translates into specific limits for the angular variables, normally with the angles ranging from 0 to ?/2, ensuring the integration covers only the desired part of the sphere.

Substitution in Radial Integration

This concept involves using substitution techniques for simplifying the integration, particularly in the radial component. When the integrand contains expressions like e^(r²) arising from an original function such as e^(x²+y²+z²), performing a substitution with u = r² can simplify the integration process. This method streamlines the calculation of the integral by reducing it to a more standard form.

Transcript

00:01 Start like that, we have x square plus y square plus z square.

00:06 It is less than equal to one.

00:08 We can rewrite this as row square less than equal to one.

00:11 So we have raw is less than equal to one, right? so our solid is a ball centered at the origin with radius three.

00:20 Now we have second restriction, right? so that is showing at the first quadrant only.

00:27 So if we just draw that in the x, y plane.

00:29 It will be first quadrant and it will be like that it will be one here this will be one here that's a restriction so we have theta lying between zero and pi by two zero less than equal to theta less than equal to pi by two and pi less than equal to five less than equal to five less than equal to five by two right so if we just sketch our ball in the in the y z plane catch our ball in the yz plane right and look at the first quadrant so get plying between 0 and pi by 2.

01:05 So we will have now the triple integral that is given as 0 to pi by 2 and then we have 0 pi by 2 and then we have 0 to 1 right we'll give so we have x will be row sine 5 cos theta we'll have e to the power row square because we had e to the power x square plus y square plus z square right that would give it to the power row square and then we have square, sine 5, d row, d5, d5, d theta, all right, just we have here.

01:43 We just keep on solving this.

01:44 We separate the integrals.

01:46 We'll have 0 5x2, cost theta, d theta, then 0 .2, pi by 2.

01:54 We'll have sine square 5, d phi.

01:58 And then we have 0 to 1...

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