Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the y-axis. $y=\frac{x^{2}}{2}+4$, $0 \leq x \leq 2$

Find the volume of the solid generated by revolving the region in the irst quadrant bounded by the coordinate axes and the curve

$$y = \cos x , 0 \leq x \leq \pi / 2$$

about

the line

$$x = \pi / 2.$$

Let C be the curve x = f(t), y = g(t), for $a \leq t \leq b$, where f' and g' are continuous on[a, b] and C does not intersect itself, except possibly at its endpoints. If g is nonnegative on [a, b], then the area of the surface obtained by revolving C about the x-axis is $s=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$. Likewise, if f is nonnegative on [a, b], then the area of the surface obtained by revolving C about the y-axis is $S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$. Find the area of the surface obtained by revolving one arch of the cycloid x = t - sin t, y = 1 - cos t, for $0 \leq t \leq 2 \pi$, about the x axis.

Calculate the area of the surface generated by revolving the parametric curve $x=\cos ^2 t, y=\sin ^2 t(0 \leq t \leq \pi / 2)$ about the y-axis.

Use the parametric equations $x=t^{2} \sqrt{3}$ and $y=3 t-\frac{1}{3} t^{3}$ to answer the following. Find the surface area generated by revolving the curve about the x-axis.

Find the surface area of the solid generated by revolving each curve about the x-axis. $x(t)=3 t^{2}$, $y(t)=6 t$; $0 \leq t \leq 1$

Let C be the curve x=f(t), y=g(t), for $a \leq t \leq b$, where f' and g' are continuous on [a. b] and C does not intersect itself, except possibly at its endpoints. If g is nonnegative on [ a, b ], then the area of the surface obtained by revolving C about the x-axis is $S=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$. Likewise, if f is nonnegative on [a, b], then the area of the surface obtained by revolving C about the y-axis is $S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$. These results can be derived. Find the area of the surface obtained by revolving the curve $x=\cos ^{3} t, y=\sin ^{3} t$, for $0 \leq t \leq \pi / 2$, about the x-axis.

Use a graphing utility to (a) plot the graphs of $y = \sin x$ and $y = x^2$ (b) find the approximate x-coordinates of the points of intersection of the graphs, and (c) find an approximation of the volume of the solid obtained by revolving the region bounded by the graphs of the functions about the y-axis.

Graph the parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range.

$$x=(y-3)^2$$

Let C be the curve x = f(t), y = g(t), for $a \leq t \leq b$, where f' and g' are continuous on[a, b] and C does not intersect itself, except possibly at its endpoints. If g is nonnegative on [a, b], then the area of the surface obtained by revolving C about the x-axis is $s=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$. Likewise, if f is nonnegative on [a, b], then the area of the surface obtained by revolving C about the y-axis is $S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$. Find the area of the surface obtained by revolving the curve $x=\cos ^{3} t, y=\sin ^{3} t$, for $0 \leq t \leq \pi / 2$, about the x-axis.

Let C be the curve x=f(t), y=g(t), for $a \leq t \leq b$, where f' and g' are continuous on [a. b] and C does not intersect itself, except possibly at its endpoints. If g is nonnegative on [ a, b ], then the area of the surface obtained by revolving C about the x-axis is $S=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$. Likewise, if f is nonnegative on [a, b], then the area of the surface obtained by revolving C about the y-axis is $S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$. These results can be derived. Find the area of the surface obtained by revolving one arch of the cycloid x=t-sin t, y=1-cos t, for $0 \leq t \leq 2 \pi$, about the x-axis.

Write in the blank whether the sentence is simple or compound.

Lauren is sweeping the dugout, and Kat is cleaning the bases.

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify your results using the integration capabilities of a graphing utility. \

$$y=\cos x, y=0, x=0, x=\dfrac{\pi}{2}$$

Find the area of the surface generated by revolving the curve about (a) the x-axis and (b) the y-axis. x = 2 cos θ, y = 2 sin θ, 0 ≤ θ ≤ π/2