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?

Question

Scarlett Allison · Accepted Answer

# 128/3# cu

#V = int int z dx dy, x >= 0, z in [ 0, x ], and x^2 + y^2 <= 16# #= int x^2/2 dy, x^2 + y^2 <= 16.# #= 1/2 int (16- y^2 ) dy. y in [ -4, 4 ]# #=1/2 [ 16 y - y^3/3 ]#, between y = -4 and y = 4 # = 128/3# cu.

Cameron Alexander · Answer

undefined 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.

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?