5 stars for correct answers 1 2 3 4 5 Using polar coordinates, evaluate the inte

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Using polar coordinates, evaluate the integral [math] where R is the region [math]. A cylindrical drill with radius 2 is used to bore a hole through the center of a sphere of radius 7. Find the volume of the ring shaped solid that remains. From Rogawski ET 2e section 15.4, exercise 21. Find the volume of the wedge shaped region (Figure 1) contained in the cylinder [math] and bounded above by the plane [math] and below by the [math]-plane. Convert the integral to polar coordinates and evaluate it (use t for theta): With a= , b = , c = and d = , dy dx = dr dt = dt = =.Consider the solid shaped like an ice cream cone that is bounded by the function z= and z = . Set up an integral in polar coordinates to find the volume of this ice cream cone. Instructions: Please enter the integral in the first answer box, typing theta for theta. Depending on the order of integration you choose, enter dr and d theta in either order into the second and third answer boxes with one dr or d theta in each box. Then, enter the limits of integration and evaluate the integral to find the volume. A = B = C = D = Volume =

Explanation / Answer

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bounded below by xy plane means z is grounded so z = 0

using cylindrical coordinates:

x = r cos(?) & y = r sin(?) & r^2 = x^2 + y^2

x^2 + y^2 = 9 --------> r^2 = 9 ------> r = +/- 3

? will be from 0 to 2? to let it rotate the whole region.


0 ? ? ? 2? ? 0 ? r ? 3 ? 0 ? z ? x ? dz r dr d?

x
? r dz
0

......x
r z ] = r * ( x - 0 ) -------> r x which is r * r cos(?) = r^2 cos(?)
...0

3
? ( r^2 cos(?) ) dr
0
. . . . . . . . . . . . . .3
cos(?) * (1/3) * r^3 ]
. . . .. . . . . . . . . 0

cos(?) * (1/3) * (3^3 - 0^2)
cos(?) * (1/3) * (3^3)
cos(?) * (3^2)
9cos(?)

2?
? 9cos(?) d? ====> it is symmetric
0

?/2
? 4 * 9cos(?) d?
0

. . . .. . .?/2
36 sin(?) ] = 36 ( sin(?/2) - sin(0) ) = 36 * (1 - 0) = 36 unit^3
. . . . . . .0

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