# How do you find the volume of the wedge-shaped region on the figure contained in the cylinder #x^2 + y^2 = 16# and bounded above by the plane #z = x# and below by the xy-plane?

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To find the volume of the wedge-shaped region, we first need to determine the limits of integration for the triple integral. The region is bounded by the cylinder (x^2 + y^2 = 16), which is a circle in the xy-plane, and the plane (z = x).

Since the region is symmetric about the z-axis, we can restrict our attention to the first octant (where (x, y, z \geq 0)). In this octant, the region is bounded by the circle (x^2 + y^2 = 16) and the plane (z = x).

To find the limits of integration, we first solve the cylinder equation for (y) in terms of (x): (y = \sqrt{16 - x^2}). Since the region is in the first octant, (y) varies from 0 to (\sqrt{16 - x^2}), (x) varies from 0 to 4 (the radius of the circle), and (z) varies from 0 to (x).

Thus, the volume of the region is given by the triple integral:

[ V = \int_0^4 \int_0^{\sqrt{16 - x^2}} \int_0^x dz , dy , dx ]

Integrating with respect to (z) first, we get:

[ V = \int_0^4 \int_0^{\sqrt{16 - x^2}} x , dy , dx ]

Integrating with respect to (y), we get:

[ V = \int_0^4 xy \Bigg|_0^{\sqrt{16 - x^2}} , dx ]

[ V = \int_0^4 x\sqrt{16 - x^2} , dx ]

This integral can be solved using a trigonometric substitution. Let (x = 4\sin(\theta)), then (dx = 4\cos(\theta) , d\theta). Substituting, we get:

[ V = \int_0^{\frac{\pi}{2}} 4\sin(\theta)\sqrt{16 - 16\sin^2(\theta)} \cdot 4\cos(\theta) , d\theta ]

[ V = \int_0^{\frac{\pi}{2}} 16\sin(\theta)\cos(\theta) \cdot 4\cos(\theta) , d\theta ]

[ V = \int_0^{\frac{\pi}{2}} 64\sin(\theta)\cos^2(\theta) , d\theta ]

Using the identity (\cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta))), we have:

[ V = \int_0^{\frac{\pi}{2}} 32\sin(\theta) + 32\sin(\theta)\cos(2\theta) , d\theta ]

[ V = 32\int_0^{\frac{\pi}{2}} \sin(\theta) , d\theta + 32\int_0^{\frac{\pi}{2}} \sin(\theta)\cos(2\theta) , d\theta ]

[ V = 32\Big[-\cos(\theta)\Big]_0^{\frac{\pi}{2}} + 32\Big[-\frac{1}{2}\cos(2\theta)\Big]_0^{\frac{\pi}{2}} ]

[ V = 32\Big(-(-1) - (-\cos(0))\Big) + 32\Big(-\frac{1}{2}\cos(\pi) - (-\frac{1}{2}\cos(0))\Big) ]

[ V = 32(1 - (-1)) + 32(-\frac{1}{2}(-1) - (-\frac{1}{2})) ]

[ V = 32(2) + 32(-\frac{1}{2} + \frac{1}{2}) ]

[ V = 64 + 0 ]

[ V = 64 ]

Therefore, the volume of the wedge-shaped region is 64 cubic units.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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