To rotate a curve around the y-axis, first convert the function so that y is the independent variable by solving  for x. This leads to the function . 

We'll also need to convert the endpoints of the interval to y-values. Note that when , and when  Therefore, the the interval being rotated is from .

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

, where V is volume,  are the endpoints of the interval, and  the function being rotated. 

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

.

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

. 

.

The formula for the volume is given as

where  and the bounds on the integral come from the interval .

As such, 

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

Applying this rule we get

And by the corollary of the First Fundamental Theorem of Calculus

As such,

 units cubed

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

where 

As such

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

Applying this rule term by term we get

And by the corollary of the Fundamental Theorem of Calculus 

As such the volume is

 units cubed

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

Using the formula for cylindrical shells, we have

.

Find the area bound by the curve of g(t), the x and y axes, and the line 

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 . These will be our limits of integration. With that in mind, our integral look like this.

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

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

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

And our upper limit...

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

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:

where ds is given by the equation below:

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:

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

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:

, so .

Both the terms of the force are power terms in the form , which have the integral , so the integral of the force is .

We know 

.

This means 

.

The length of the curve of  on the interval  can be determined by evaluating the integral

.

so

 .

The above integral becomes 

Substitute . Then , , and the integral becomes

The length of the curve of  on the interval  can be determined by evaluating the integral

.

, so 

The integral becomes

Use substitution - set . Then , and . The bounds of integration become  and , and the integral becomes

The general formula for finding the length of a curve  over an interval  is 

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

.

The derivative of  can be found using the power rule, , which leads to 

. 

At this point, a substitution is useful.

Let 

.

We can also express the limits of integration in terms of  to simplify computation. When , and when .

Making these substitutions leads to 

.

Now use the power rule, which in general is , to evaluate the integral.