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1.11738 NIntegrate[x^2 + y^2, {x, 0.724492, 1.22074}, {y, x^2, Sqrt[x + 1]}] 1.11738
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Multivariate Calculus
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10.20 Find the volume under the hemisphere z = 4 x2 y2 above the region in the x-y plane bounded by the cardioid r = 1 cos q.
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We will translate the problem into cylindrical coordinates. Since r2 = x2 + y2, the equation of the hemisphere becomes z = 4 r2. The region of integration, R, is the cardioid shown. PolarPlot[1 Cos[p], {p, 0, 2o}]
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The volume, V =
(4 r
) dA =
1 cos
(4 r 2 ) r dr d
1-Cos[ ] 0
(4 r2)r p r
61 16
10.21 Find the volume of the solid that lies under the paraboloid z = x2 + y2, above the x-y plane, and inside the cylinder (x 1)2 + y2 = 1.
SOLUTION
The cylinder (x 1)2 + y2 = 1 is a cylinder of radius 1 whose axis is translated from the z-axis by the vector (1, 0, 0).
s1 = Graphics3D[Cylinder[{{1, 0, 0}, {1, 0, 8}}]]; s2 = Plot3D[x2 + y2, {x, 2, 2}, {y, 2, 2}];
Multivariate Calculus
Show[s1, s2, PlotRange {0,4}, ViewPoint {1.217, 3.125, 0.447}]
Now that we know what the region looks like, we must look at its projection in the x-y plane. ContourPlot[(x 1)2 + y2 1, {x, 0, 2},{y, 1, 1},Frame False, Axes True]
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Although the problem can be solved in rectangular coordinates, it is easier to solve it using cylindrical coordinates. The equation of the circle can be expanded to x2 + y2 = 2x, which is equivalent, in polar coordinates, to r = 2 cos q, and the paraboloid z = x2 + y2 becomes z = r2. The complete circle is generated as q varies from p/2 to p/2. The required volume may be expressed as a double integral:
3 2
2Cos[ ]
/ 2
/2
2 cos
(r 2 ) r dr d
/2 0
r r 3 p
10.22 The area of the surface z = f(x, y) above the region R in the x y plane is Compute the surface area of a sphere of radius a.
SOLUTION
[ fx ( x , y)] 2 + [ f y ( x , y)] 2 + 1.
We compute the surface area of the portion of the sphere in the first octant and, by symmetry, multiply by 8. The equation of the sphere is x2 + y2 + z2 = a2. Solving for z we get z = f ( x , y) = a 2 x 2 y 2 as the function representing the upper hemisphere. Since the projection of the hemisphere onto the x y plane is a circle of radius a centered at the origin, it is most convenient to use cylindrical coordinates.
Multivariate Calculus
f[x_, y_]= a2 x2 y 2 ;
2 2 1 +( x f[x, y]) +( y f[x, y]) // Together
a2 -a + x2 + y 2
% /. x2 + y 2 r2 // Simplify a2 a2 r2
Replace x 2 + y 2 by r 2 .
1 + ( x f [ x, y]) 2 + ( y f [ x, y]) 2 dA =
0 0
/2 a
a2 a r2
r p r
4 a2 Abs[a] Simplify[%, Assumptions a > 0] 4 a2 p
/2
a2 r dr d a2 r 2
10.23 Find the volume of the ice cream cone bounded by the cone z = 3 x 2 + y 2 and the sphere
x 2 + y 2 + (z 9)2 = 9.
SOLUTION
The required volume is represented by the triple integral
dV .
Because of the nature of the bounding
surfaces, this problem is done most conveniently using cylindrical coordinates. First, rewrite the equation of the sphere, solving for z in terms of x and y. Solve[x2 + y2 + (z 9)2 9, z]
{{z 9
9 x2 y 2 , z 9 + 9 x2 y 2
Using the second solution (corresponding to the upper hemisphere), and replacing x2 + y2 by r2, the equation of the sphere becomes z = 9 + 9 r 2 . The equation of the cone, z = 3 x 2 + y 2 , becomes z = 3r. Now we can sketch the surfaces that form our region.
cone = RevolutionPlot3D[3 r,{r, 0, 3},{p, 0, 2 o}];
hemisphere = RevolutionPlot3D 9 + 9 r2 ,{r, 0,3},{p,0,2o } ;
Show [cone, hemisphere, PlotRange All, BoxRatios {1, 1, 2}, Axes False, Boxed False]
Multivariate Calculus
To compute the volume, we observe that the projection of the region onto the x-y plane is a circle. To determine its radius, we find the intersection of the cone and the hemisphere. Solve 3 r 9 + 9 r2 {{r 3}} The projection onto the x-y plane is a circle of radius 3 centered at the origin. The required volume is
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