Double integrals—your choice of transformation Evaluate the following integrals using a change of variables. Sketch the original and new regions of integration , R and S. 32. ∬ R y 2 − x 2 d A , where R is the diamond bounded by y – x = 0, y – x = 2, y + x = 0, and y + x = 2
Double integrals—your choice of transformation Evaluate the following integrals using a change of variables. Sketch the original and new regions of integration , R and S. 32. ∬ R y 2 − x 2 d A , where R is the diamond bounded by y – x = 0, y – x = 2, y + x = 0, and y + x = 2
Solution Summary: The author evaluates the value of the integral and sketches the original and new region.
Double integrals—your choice of transformationEvaluate the following integrals using a change of variables. Sketch the original and new regions of integration, R and S.
32.
∬
R
y
2
−
x
2
d
A
, where R is the diamond bounded by y – x = 0, y – x = 2, y + x = 0, and y + x = 2
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
According to Newton's law of universal gravitation, the force F between two bodies of constant mass
GmM
m and M is given by the formula F =
, where G is the gravitational constant and d is the
d²
distance between the bodies.
a. Suppose that G, m, and M are constants. Find the rate of change of force F with respect to
distance d.
F' (d)
2GmM
b. Find the rate of change of force F with gravitational constant G = 6.67 × 10-¹¹ Nm²/kg², on
two bodies 5 meters apart, each with a mass of 250 kilograms. Answer in scientific notation,
rounding to 2 decimal places.
-6.67x10
N/m syntax incomplete.
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