88. The region in Exercise 87 is revolved about the x-axis to ...

Question

88. The region in Exercise 87 is revolved about the x-axis to generate a solid.

b. Show that the inner and outer surfaces of the solid have infinite area.

Accepted Answer

Consider the region in the first quadrant bounded above by Y equals co sequin x, below by Y equals cotangent x, and between the vertical lines x equals 0 and X equals pi divided by 2. Do the inner and outer surfaces of the solid formed when this region is evolved about the x axis have infinite area? Yes or no. Now, let's first look at the surface area formula, which is given by S equals 2 pi. Multiplied by the integral for a the b. Of F of X. Square root 1 + F of X. Squared D X Now, we do have two surfaces here. F X equals coequin x. G X equals cotangent x. Let's find the derivatives. The derivative of cosequent is going to be negative cosequent x cotangent x. And the derivative. Of cotangent x is negative cosequent squared x. Now we can write our integrals down. We have our first surface integral, which is 2 pi integral. From 0 to pi divided by 2. Of cosequin X square root 1 + negative cosequin X cotangent X, all square. D X, and we have our 2nd integral. 2 pi integral from 0 to pi divided by 2. Of cotangent X square root 1 + negative cosequent X2. All squared Dix Thus consider the limits as x approaches 0. Now, based on our functions, We have co sequin x. We'll say the limit as x approaches 0 from the right. That is the limit as x approaches 0 from the right of 1 divided by sin x. Now, sin of 0 is 0, which means we have 1 divided by 0 or infinity. We can also check the same for cotangent. Cotangent is equivalent to 1 divided by tangent X, which 1 divided by tangent of 0 is 1 divided by 0, or infinity. So then we know. From the right, both functions approach infinity. So We can now approximate our integrals. For S1 If we assume then that the approximation of co sequin x is 1 divided by x and cotangent x also 1 divided by X, we have 1 divided by x square root. of one. Plus 1 divided by x2 multiplied by 1 divided by x2. Which is 1 divided by X square root. Of 1 + 1 divided by x to the 4th. Meaning these both approach approximately. 1 divided by x to the 3rd. So Both of these integrals have anagrams that tend to infinity if we would repeat this for S2. So from here, We can then assume that we have the integral. From 0 to some infinitely small value epsilon. Of 1 divide by x to the 3rd dx, we can use as approximation instead of actually solving the integrals themselves. Now this is an improper integral. We have the limit. As T approaches 0 from the right. Of the integral from T to epsilon. X 3 D X. This is the same as saying the limit ast approaches zero from the right. Of one divided by 2 x 2 with a negative in the front. From T to epsilon. We get the limit Ast approaches 0 from the right. Of -1 divided by 2 epsilon 2. Plus 1 divided by 2 t 2. Where once again, epsilon is some small value that is greater than 0. Now, based on everything we said here with our limits, this is going to approach infinity. Which means both of our integrals should diverge. If we then say that both of them diverge, we have infinite area, which means our answer is yes. When revolved about the x axis, we have infinite area. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

88. The region in Exercise 87 is revolved about the x-axis to generate a solid.b. Show that the inner and outer surfaces of the solid have infinite area.

Watch next

88. The region in Exercise 87 is revolved about the x-axis to generate a solid.
b. Show that the inner and outer surfaces of the solid have infinite area.