### Video Transcript

Find the area of the region bounded by the polar curve π is equal to the square root of the natural logarithm of π, where π is greater than or equal to one and π is less than or equal to two π.

The question wants us to find the area of a region which is bounded by a polar curve where π is greater than or equal to one and π is less than or equal to two π. To start, we know how to find the area of a region bounded by a polar curve. We know if we have a polar curve π, which is equal to some function π of π, where π is continuous and nonnegative on the closed interval from π one to π two, then we can find the area of the region bounded by the curve π is equal to π of π where π is between π one and π two. This area is equal to the integral from π one to π two of one-half times π of π squared with respect to π.

And itβs worth pointing out since we have π is equal to π of π, sometimes we write our integrand as one-half times π squared. To start, since we want to find the area of the region bounded by the polar curve π is equal to the square root of the natural logarithm of π, weβll set our function π of π to be the square root of the natural logarithm of π. Next, weβll set our values of π one and π two to be the endpoints of our interval. Thatβs π one is one and π two is to π.

So to use our formula, we need to check that our function π is continuous and nonnegative for all values of π between one and two π. Letβs start by checking that π is a continuous function. We can see that π is the square root of the natural logarithm of π. And we know the natural logarithm function is continuous and the square root function is continuous. So π of π is the composition of two continuous functions. And we know the composition of continuous functions will be continuous on its domain. So we just need to find the domain of our function π of π.

So letβs find the domain of our function π of π. First, we know we canβt take the natural logarithm of any number less than or equal to zero. So this tells us that our values of π must be positive. Next, remember weβre taking the square root of this value. And we canβt take the square root of a negative number. So this tells us for π to be in the domain of our function π, we must have the natural logarithm of π be greater than or equal to zero. Otherwise, weβre taking the square root of a negative number.

And remember, the natural logarithm function is an increasing function. And we know the natural logarithm of one is equal to zero. This tells us that π must be greater than or equal to one. So the domain of π of π is when π is greater than zero and π is greater than or equal to one. Of course, we can simplify this to just be the values of π greater than or equal to one. And in particular, this means we have shown that our function π of π is continuous on our closed interval from one to two π.

Now all we need to do is show that our function is nonnegative on this interval. If we look at our function π of π, we can see that the last thing we do is take the positive square root of a number. So because the interval from one to two π is strictly within the domain of our function π of π, all of our outputs will be nonnegative. So we can now use this formula to find the area of the region bounded in our question. So by using this formula and substituting in our values for π one, π two, and π of π, we get that the area π΄ is equal to the integral from one to two π of one-half times the square root of the natural logarithm of π all squared with respect to π.

And of course, we can simplify this expression. The square root of the natural logarithm of π all squared is just equal to the natural logarithm of π. So we just need to evaluate the integral from one to two π of one-half times the natural logarithm of π with respect to π. The easiest way to evaluate integrals only involving the natural logarithm function is to use integration by parts. We recall integration by parts tells us the integral from π to π of π’ multiplied by π£ prime with respect to π is equal to π’ times π£ evaluated at the limits of integration π and π minus the integral from π to π of π’ prime π£ with respect to π.

We have several different methods of choosing which of the factors of our integrand will be π’ and which one will be π£ prime. However, in this case, that doesnβt seem to be much choice. The natural logarithm function is very difficult to integrate. However, itβs easy to differentiate. So weβll set our function π’ to be the natural logarithm of π and π£ prime to be one-half. Now, to use integration by parts, we need to find expressions for π’ prime and π£. Weβll start with π’ prime. Thatβs the derivative of the natural logarithm of π with respect to π. We know this is equal to one divided by π.

Next, we need to find an expression for π£. π£ is the antiderivative of one-half, which we know is π divided by two. Of course, we couldβve added a constant. However, weβre working with a definite integral, so we donβt need to. Weβre now ready to evaluate this integral by using integration by parts. We get π’ times π£ evaluated at the limits of integration one and two π minus the integral from one to two π of π’ prime π£ with respect to π. Substituting in our expressions for π’, π’ prime, and π£, we get the natural logarithm of π multiplied by π over two evaluated at the limits of integration one and two π minus the integral from one to two π of one over π times π over two with respect to π.

Before we start evaluating this expression, we can simplify. In our integrand, we have π divided by π, so we can cancel this to get a new integrand of one-half. Next, to simplify this expression slightly, weβll take out the factor of one-half from our first term. Weβre now ready to evaluate this expression. Letβs start by evaluating the limits of integration in our first term. Evaluating this at the limits of integration, π is equal to one and π is equal to two π, we get one-half times the natural logarithm of two π multiplied by two π minus the natural logarithm of one times one. And of course, we know the natural logarithm of one is equal to zero. So we can just cancel this part of the expression out. And we can still simplify this. One-half multiplied by two is equal to one.

Letβs now evaluate the integral in our second term. First, by using the power rule of integration, the integral of one-half with respect to π is π over two. And all thatβs left now is to evaluate this at the limits of integration. We get two π over two minus one-half. And of course we can simplify this. Two π divided by two is just equal to π. And finally, weβll distribute the negative over our parentheses. And this gives us our final answer.

Therefore, we were able to show the area of the region bounded by the polar curve π is equal to the square root of the natural logarithm of π, where π is greater than or equal to one and π is less than or equal to two π, is equal to π times the natural logarithm of two π minus π plus one-half.