To rotate a curve around the y-axis, first convert the function so that y is the independent variable by solving \(\displaystyle y=\sqrt{x}\) for x. This leads to the function \(\displaystyle x=y^2\). 

We'll also need to convert the endpoints of the interval to y-values. Note that when \(\displaystyle x=4, \ y=2\), and when \(\displaystyle x = 9, y = 3.\) Therefore, the the interval being rotated is from \(\displaystyle y=2\ to\ y=3\).

The disk method is best in this case. The general formula for the disk method is 

\(\displaystyle V=\pi\int_{a}^{b}f^2(x)dx\), where V is volume, \(\displaystyle a\ and\ b\) are the endpoints of the interval, and \(\displaystyle f(x)\) the function being rotated. 

Substuting the function and endpoints from the problem at hand leads to the integral

\(\displaystyle V=\pi\int_{2}^{3}(y^2)^2dy=\pi\int_{2}^{3}y^4dy\).

To evaluate this integral, you must know the power rule. Recall that the power rule is 

\(\displaystyle \int{y^ndy}=\frac{y^{n+1}}{n+1}+c\). 

\(\displaystyle \pi\int_{2}^{3}y^4dy=\pi[\frac{y^5}{5}]_{2}^{3}\)

\(\displaystyle =\pi\left[\frac{3^5}{5}-\frac{2^5}{5}\right]=\pi\left[\frac{243}{5}-\frac{32}{5}\right]\)

\(\displaystyle =\frac{211\pi}{5}\).

The formula for the volume is given as

\(\displaystyle V=\pi \int_{a}^{b} r^2 \, dx\)

where \(\displaystyle r=f(x)\) and the bounds on the integral come from the interval \(\displaystyle [a,b]\).

As such, 

\(\displaystyle V=\pi \int_{0}^{8} x^2 \, dx\)

When taking the integral, we will use the inverse power rule which states

\(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\)

Applying this rule we get

\(\displaystyle \pi \int_{0}^{8} x^2 \, dx=\pi \left(\frac{x^3}{3}\right) \left.\begin{matrix} \\ \end{matrix}\right|_{0}^{8}\)

And by the corollary of the First Fundamental Theorem of Calculus

\(\displaystyle \pi \left(\frac{x^3}{3}\right) \left.\begin{matrix} \\ \end{matrix}\right|_{0}^{8}=\pi \left(\left[\frac{8^3}{3}\right]-\left[\frac{0^3}{3}\right]\right)\)

\(\displaystyle =\frac{512\pi}{3}\)

As such,

\(\displaystyle V=\frac{512\pi}{3}\) units cubed

The formula for volume of the region revolved around the x-axis is given as 

\(\displaystyle V=\pi \int_{a}^{b} r^2\, dx\)

where \(\displaystyle r=f(x)\)

As such

\(\displaystyle V=\pi \int_{0}^{10} (-x^2+10x+3)^2\,dx = \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx\)

When taking the integral, we use the inverse power rule which states

\(\displaystyle \int x^n\,dx = \frac{x^{n+1}}{n+1}\)

Applying this rule term by term we get

\(\displaystyle = \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx = \pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}\)

And by the corollary of the Fundamental Theorem of Calculus 

\(\displaystyle \pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}=\pi \left ( \left [ \frac{(10)^5}{5}-\frac{20(10)^4}{4}+\frac{94(10)^3}{3}+\frac{60(10)^2}{2}+9(10) \right ]-0 \right )\)

\(\displaystyle =\frac{13270}{3}\pi\)

As such the volume is

\(\displaystyle \frac{13270}{3}\pi\) units cubed

Since we are revolving a function of \(\displaystyle x\) around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

\(\displaystyle \int_{0}^{4}2\pi x (\sqrt{x})dx\)

\(\displaystyle =\int_{0}^{4}2\pi x^{3/2}dx\)

\(\displaystyle =[2\pi x^{5/2}]^{4}_{0}\)

\(\displaystyle =\frac{128}{5}\pi\).

Find the area bound by the curve of g(t), the x and y axes, and the line \(\displaystyle t=\frac{3 \pi}{4}\)

\(\displaystyle g(t)=4sin(t)-6cos(t)\)

We are asked to find the area under a curve. This sounds like a job for an integral.

We need to integrate our function from 0 to \(\displaystyle t=\frac{3 \pi}{4}\). These will be our limits of integration. With that in mind, our integral look like this.

\(\displaystyle G(t)=\int_0^\frac{3\pi}{4}4sin(t)-6cos(t)dt\)

Now, we need to recall the rules for integrating sine and cosine.

\(\displaystyle 1) \int sin(t)dt=-cos(t)+c\)

\(\displaystyle 2) \int cos(t)dt=sin(t)+c\)

With these rules in mind, we can integrate our original function to get:

\(\displaystyle G(t)=\int_0^\frac{3\pi}{4}4sin(t)-6cos(t)dt=-4cos(t)-6sin(t)+c|_0^{\frac{3 \pi}{4}}\)

\(\displaystyle G(t)=-4cos(t)-6sin(t)+c|_0^{\frac{3 \pi}{4}}\)

Now, we just need to evaluate our new function at the given limits. Let's start with 0

\(\displaystyle G(0)=-4cos(0)-6sin(0)+c=-4-0+c=-4+c\)

And our upper limit...

\(\displaystyle G({\frac{3 \pi}{4}})=-4cos({\frac{3 \pi}{4}})-6sin({\frac{3 \pi}{4}})+c\)

\(\displaystyle -4cos({\frac{3 \pi}{4}})-6sin({\frac{3 \pi}{4}})+c=(-4)\frac{-\sqrt{2}}{2}-(6)\frac{-\sqrt{2}}{2}+c\)

\(\displaystyle (-4)\frac{-\sqrt{2}}{2}-(6)\frac{\sqrt{2}}{2}+c=2\sqrt{2}-3\sqrt{2}+c=-\sqrt{2}+c\)

Now, we need to take the difference between our limits:

\(\displaystyle G(\frac{3 \pi}{4})-G(0)=(-\sqrt{2}+c)-(-4+c)=-\sqrt{2}+4\approx2.585\)

So, our answer is 2.59

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

\(\displaystyle L=\int ds\)

where ds is given by the equation below:

 \(\displaystyle ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx\)

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

\(\displaystyle \frac{dv}{dt}=(t-1)^{\frac{1}{2}}=\sqrt{t-1}\)

Now we can plug this into the given equation to find ds:

\(\displaystyle ds=\sqrt{1+(\frac{dv}{dt})^{2}}dt=\sqrt{1+(\sqrt{t-1})^{2}}=\sqrt{1+t-1}=\sqrt{t}\)

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

\(\displaystyle L=\int ds=\int_{1}^{4}\sqrt{t}dt=[\frac{2}{3}t^{\frac{3}{2}}]_{1}^{4}\)

\(\displaystyle =\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}\)

\(\displaystyle \small \small W=\int (Fdx)\), so \(\displaystyle \small W=\int(3x^{2}+x)dx\).

Both the terms of the force are power terms in the form \(\displaystyle \small ax^{n}\), which have the integral \(\displaystyle \small \frac{a}{n+1}x^{n+1}\), so the integral of the force is \(\displaystyle \small x^{3}+x^{2}/2+c\).

We know 

\(\displaystyle \small W(0)=0\rightarrow 0^{3}+0^{2}/2+c=0 \therefore c=0\).

This means 

\(\displaystyle \small W(x)=x^{3}+x^{2}/2\).

The length of the curve of \(\displaystyle f(x)\) on the interval \(\displaystyle [a,b]\) can be determined by evaluating the integral

\(\displaystyle \int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x\).

\(\displaystyle f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}\)

so

\(\displaystyle f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}\) .

The above integral becomes 

\(\displaystyle \int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} \; \mathrm{d}x\)

\(\displaystyle \int_{0}^{3} \sqrt{x+1} \; \mathrm{d}x\)

Substitute \(\displaystyle u = x + 1\). Then \(\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = 1\), \(\displaystyle \mathrm{d} u = \mathrm{d} x\), and the integral becomes

\(\displaystyle \int_{1}^{4} \sqrt{u} \; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} \; \mathrm{d}u\)

\(\displaystyle = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}\)

\(\displaystyle = \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}\)

\(\displaystyle = \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}\)

\(\displaystyle = \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}\)

The length of the curve of \(\displaystyle f(x)\) on the interval \(\displaystyle [a,b]\) can be determined by evaluating the integral

\(\displaystyle \int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x\).

\(\displaystyle f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )\), so 

\(\displaystyle f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x\)

The integral becomes

\(\displaystyle \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{ \frac{\pi}{8}} \sec 2x \; \mathrm{d}x\)

Use substitution - set \(\displaystyle u = 2x\). Then \(\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = 2\), and \(\displaystyle \mathrm{d} x = \frac{1}{2}\mathrm{d} u\). The bounds of integration become \(\displaystyle 2 \cdot 0 = 0\) and \(\displaystyle 2 \cdot \frac{\pi}{8} = \frac{\pi}{4}\), and the integral becomes

\(\displaystyle \frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u \; \mathrm{d}u\)

\(\displaystyle =\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \\ \\ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\\ \\ 0 \end{matrix}\)

\(\displaystyle =\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |\)

\(\displaystyle =\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |\)

\(\displaystyle =\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1\)

\(\displaystyle =\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0\)

\(\displaystyle =\frac{1}{2} \ln \left (1 + \sqrt{2} \right )\)

The general formula for finding the length of a curve \(\displaystyle f(x)\) over an interval \(\displaystyle [a,b]\) is \(\displaystyle \int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx\)

In this example, the arc length can be found by computing the integral

\(\displaystyle \int_{0}^{4}\sqrt{1+(\frac{d}{dx}x^{\frac{3}{2}})^2}\ dx\).

The derivative of \(\displaystyle x^{\frac{3}{2}}\) can be found using the power rule, \(\displaystyle \frac{d}{dx} x^n=nx^{n-1}\), which leads to 

\(\displaystyle \int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}\ dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx\). 

At this point, a substitution is useful.

Let 

\(\displaystyle u=1+\frac{9}{4}x,\ du=\frac{9}{4}dx,\frac{4}{9}du=dx\).

We can also express the limits of integration in terms of \(\displaystyle u\) to simplify computation. When \(\displaystyle x=0, u=1\), and when \(\displaystyle x = 4, u=10\).

Making these substitutions leads to 

\(\displaystyle \int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx=\int_{1}^{10}\frac{4}{9}\sqrt{u}du=\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du\).

Now use the power rule, which in general is \(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}+c\), to evaluate the integral. 

\(\displaystyle \frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du=\frac{4}{9}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}|_{1}^{10}\)

\(\displaystyle =\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)\)