Video: Finding the Area of a Region Bounded by a Given Curve

Find the area of the region bounded by the polar curve π‘Ÿ = 𝑒^(βˆ’1/4 πœƒ), where πœ‹/2 ≀ πœƒ ≀ πœ‹.

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Video Transcript

Find the area of the region bounded by the polar curve π‘Ÿ is equal to 𝑒 to the power of negative one-quarter πœƒ, where πœƒ is greater than or equal to πœ‹ by two and πœƒ is less than or equal to πœ‹.

The question is asking us to find the area of a region bounded by a polar curve where πœƒ is between πœ‹ by two and πœ‹. And we know a formula for finding the area bounded by a polar curve. We know if π‘Ÿ is equal to some function 𝑓 of πœƒ where 𝑓 is continuous and nonnegative on a closed interval from πœƒ one to πœƒ two. Then we can find the area bounded by the polar curve π‘Ÿ is equal to 𝑓 of πœƒ, where πœƒ is between πœƒ one and πœƒ two, by calculating the integral from πœƒ one to πœƒ two of one-half times 𝑓 of πœƒ all squared with respect to πœƒ. And it’s worth pointing out since π‘Ÿ is equal to 𝑓 of πœƒ, sometimes we’ll write our integrand as one-half times π‘Ÿ squared.

In our case, since we want the area of the polar curve π‘Ÿ is equal to 𝑒 to the power of negative one-quarter πœƒ, we’ll set our function 𝑓 of πœƒ to be 𝑒 to the power of negative one-quarter πœƒ. And we’ll set πœƒ one and πœƒ two to be the bounds of our integral. That’s πœƒ one is equal to πœ‹ by two, and πœƒ two is equal to πœ‹. So to use this formula, we’re going to need to show our function 𝑓 is continuous on the closed interval from πœ‹ by two to πœ‹ and show that our function is nonnegative on this interval. Let’s start by showing that our function is continuous.

We know that the exponential function is continuous, and multiplying by negative one-quarter is continuous. So our function 𝑓 of πœƒ is the composition of continuous functions. This means it’s continuous across its entire domain. And 𝑓 of πœƒ is certainly defined for all values of πœƒ on the closed interval from πœ‹ by two to πœ‹. So 𝑓 of πœƒ is continuous on our closed interval. Let’s now check that our function is nonnegative on this closed interval. To see that this is true, we’ll start by using our laws of exponents to rewrite our function 𝑓 of πœƒ as 𝑒 to the power of negative one-quarter all raised to the power of πœƒ. We know that 𝑒 to the power of negative one-quarter is positive. In fact, it’s approximately equal to 0.78.

So our function is some positive number raised to the power of πœƒ. We know this is positive for all possible values of πœƒ. So in particular, this tells us our function 𝑓 of πœƒ is nonnegative on this interval. So we’ve shown all of our prerequisites for the formula are true. This means we can calculate the area of the region given to us in the question by calculating the integral from πœ‹ by two to πœ‹ of one-half times 𝑒 to the power of negative one-quarter πœƒ all squared with respect to πœƒ. So what we need to do now is evaluate this definite integral. We’ll simplify this by using our laws of exponents. Instead of squaring 𝑒 to the power of negative one-quarter πœƒ, we’ll multiply our exponent by two.

This gives us the integral from πœ‹ by two to πœ‹ of one-half times 𝑒 to the power of negative one-half πœƒ with respect to πœƒ. And now we’re ready to just evaluate this integral. We know for constant π‘Ž not equal to zero, the integral of 𝑒 to the power of π‘Žπ‘₯ with respect to π‘₯ is equal to one over π‘Ž times 𝑒 to the power of π‘Žπ‘₯ plus a constant of integration 𝐢. It’s true for any variable, so we’ll call our variable πœƒ. In our case, our value of the constant π‘Ž is negative one-half. So we need to divide our coefficient of one-half by negative one-half. And of course, we know one-half divided by negative one-half is equal to negative one. So our integral simplifies to give us negative 𝑒 to the power of negative one-half πœƒ evaluated at the limits of integration, πœƒ is equal to πœ‹ by two and πœƒ is equal to πœ‹.

All that’s left to do now is evaluate this at the limits of integration. We get negative 𝑒 to the power of negative one-half πœ‹ plus 𝑒 to the power of negative one-half times πœ‹ by two. Finally, we’ll simplify our exponents and reorder our two terms to get 𝑒 to the power of negative πœ‹ by four minus 𝑒 to the power of negative πœ‹ by two. And this is our final answer. Therefore, we were able to show the area of the region bounded by the polar curve π‘Ÿ is equal to 𝑒 to the power of negative one-quarter πœƒ, where πœƒ is greater than or equal to πœ‹ by two and πœƒ is less than or equal to πœ‹. Is equal to 𝑒 to of the power of negative πœ‹ by four minus 𝑒 to the power of negative πœ‹ by two.

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