If V is the hemispherical region enclosed between the surface x^2+y^2+z^2=a^2 and the plane z=0 show tht triple integral of (x+y+z) dx dy dz =πa^4÷4
Answers
Answer:
Given a parameterization of surface \text{r}\left(u,v\right)=〈x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)〉, the parameter domain of the parameterization is the set of points in the uv-plane that can be substituted into r.
Parameterizing a Cylinder
Describe surface S parameterized by
\text{r}\left(u,v\right)=〈\text{cos}\phantom{\rule{0.2em}{0ex}}u,\text{sin}\phantom{\rule{0.2em}{0ex}}u,v〉,\text{−}\infty <u<\infty ,\text{−}\infty <v<\infty .
To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of {ℝ}^{2}, we can choose any value for u and v and plot the corresponding point. If u=v=0, then \text{r}\left(0,0\right)=〈1,0,0〉, so point (1, 0, 0) is on S. Similarly, points \text{r}\left(\pi ,2\right)=\left(-1,0,2\right) and \text{r}\left(\frac{\pi }{2},4\right)=\left(0,1,4\right) are on S.
Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. To visualize S, we visualize two families of curves that lie on S. In the first family of curves we hold u constant; in the second family of curves we hold v constant. This allows us to build a “skeleton” of the surface, thereby getting an idea of its shape.
First, suppose that u is a constant K. Then the curve traced out by the parameterization is 〈\text{cos}\phantom{\rule{0.2em}{0ex}}K,\text{sin}\phantom{\rule{0.2em}{0ex}}K,v〉, which gives a vertical line that goes through point \left(\text{cos}\phantom{\rule{0.2em}{0ex}}K,\text{sin}\phantom{\rule{0.2em}{0ex}}K,v\right) in the xy-plane.
Now suppose that v is a constant K. Then the curve traced out by the parameterization is 〈\text{cos}\phantom{\rule{0.2em}{0ex}}u,\text{sin}\phantom{\rule{0.2em}{0ex}}u,K〉, which gives a circle in plane z=K with radius 1 and center (0, 0, K).
If u is held constant, then we get vertical lines; if v is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Therefore the surface traced out by the parameterization is cylinder {x}^{2}+{y}^{2}=1 ((Figure)).
(a) Lines 〈\text{cos}\phantom{\rule{0.2em}{0ex}}K,\text{sin}\phantom{\rule{0.2em}{0ex}}K,v〉 for K=0,\frac{\pi }{2},\pi ,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{3\pi }{2}. (b) Circles 〈\text{cos}\phantom{\rule{0.2em}{0ex}}u,\text{sin}\phantom{\rule{0.2em}{0ex}}u,K〉 for K=-2,-1,1,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}2. (c) The lines and circles together. As u and v vary, they describe a cylinder.
Three diagrams in three dimensions. The first shows vertical lines around the origin. The second shows parallel circles all with center at the origin and radius of 1. The third shows the lines and circle. Together, they form the skeleton of a cylinder.
Notice that if x=\text{cos}\phantom{\rule{0.2em}{0ex}}u and y=\text{sin}\phantom{\rule{0.2em}{0ex}}u, then {x}^{2}+{y}^{2}=1, so points from S do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle 〈\text{cos}\phantom{\rule{0.2em}{0ex}}u,\text{sin}\phantom{\rule{0.2em}{0ex}}u,k〉 for some k, and therefore each point on the cylinder is contained in the parameterized surface ((Figure)).
Cylinder {x}^{2}+{y}^{2}={r}^{2} has parameterization \text{r}\left(u,v\right)=〈r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}u,r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}u,v〉, 0\le u\le 2\pi ,\text{−}\infty <v<\infty .
An image of a vertical cylinder in three dimensions with the center of its circular base located on the z axis.
Analysis
Notice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to 0\le u\le \pi ,0<v<6, then the surface would be a half-cylinder of height 6.