Question

Find the volume of the solid that is bounded above by the plane $z=4$ and bounded below by $z=\sqrt{x^{2}+y^{2}}$. (b) ( 6 points) Find the area of the region in the first quadrant that lies within the cardioid $r=1-\cos \theta$. (c) ( 3 points) of the following vector valued functions, which ones trace out the part of the parabola $y=x^{2}$ from $(-2,4)$ to $(2,4)$ as $t$ ranges over the given domain? Explain your answer. (i) $\overrightarrow{\mathrm{r}}_{1}(t)=\left\langle e^{t}, e^{2 t}\right\rangle$, $-\ln (2) \leq t \leq \ln (2)$ (ii) $\vec{r}_{2}(t)=\left\langle t^{2}, t^{4}\right\rangle$, $-\sqrt{2} \leq t \leq \sqrt{2}$ (iii) $\overrightarrow{\mathbf{r}}_{3}(t)=\left\langle 2 \sin t, 4(\sin t)^{2}\right\rangle$, $-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$ (d) (3 points) Give an example of a function $w=f(x, y, z)$ whose level surfaces at height $c$ are spheres of radius $c^{2}$ for all $c \in \mathbb {R}, c>0$. CS.JG. 135

Answer 1

To find the volume of the solid bounded above by the plane z=4 and bounded below by z=√(x^2 + y^2), set up a triple integral over the region. Integrate z, r, and theta to calculate the volume. The volume is 32π.

Step-by-step explanation:

To find the volume of the solid bounded above by the plane z=4 and bounded below by z=√(x^2 + y^2), we need to set up a triple integral over the region. In cylindrical coordinates, the region can be described as 0 <= z <= 4 and 0 <= r <= 4, where r is the radius of the circular cross section.

To set up the triple integral, we integrate z from 0 to 4, r from 0 to 4, and theta from 0 to 2π. The integrand is 1, since we are just calculating the volume.

V = ∫∫∫ 1 dz dr dθ = ∫∫ [4r] dr dθ = ∫ [2π * (4^2)] dθ = 32π.