Question

Approximation \(\quad\) In Exercises 41 and 42, determine which value best approximates the volume of the solid between the \(x y\) -plane and the function over the region. (Make your selection on the basis of a sketch of the solid and not by performing any calculations.) \(f(x, y)=15-2 y ; R:\) semicircle: \(x^{2}+y^{2}=16, y \geq 0\) (a) 100 (b) 200 (c) 300 (d) -200 (e) 800

Short answer

The value that best approximates the volume of the solid is (a) 100.

Step-by-step solution

Visualize the region and the function

First, visualize the region defined by the semicircle \(x^{2}+y^{2}=16, y \geq 0\). It's basically a semicircle in the upper half of the xy-plane with radius 4. Then, visualize the function \(f(x, y)=15-2 y\). It's a plane that slopes downwards as y increases and it's always above the xy-plane as \(y >= 0\).

Visualize the solid

Now, visualize the solid between the xy-plane and the function over the defined region. It's a hemispherical cap-like shape, getting smaller as y increases.

Compare with the known volume of a hemisphere

To approximate the volume of the solid, compare it with the known volume of a hemisphere with radius 4. The volume of such a hemisphere is \(V=\frac{2}{3}\pi r^3 = \frac{2}{3}\pi (4)^3 = 134\) cubic units approximately.

Approximate the volume

However, the solid is not exactly a hemisphere such as it's somewhat smaller due to the contribution of the \(2y\) term in the function. This causes the function to decrease as \(y\) increases. Thus, the volume of the solid would be less than 134 cubic units. The closest positive value among the options is 100, so this is the best approximation.