90. Consider the infinite region in the first quadrant bounded...

Question

90. Consider the infinite region in the first quadrant bounded by the graphs of

y = 1 / √x, y = 0, x = 0, and x = 1.

b. Find the volume of the solid formed by revolving the region (i) about the x-axis

Accepted Answer

Let R be the infinite region in the first quadrant bounded by the curves, Y equals 1 divided by x 2/3. Y equals 0, X equals 0, and X equals 8. Find the volume of the solid obtained by revolving r about the x axis. We have 4 possible answerspi divided by 8 cubic units, pi divided by 43 pi divided by 4, both in cubic units, or d, the volume integral diverges. Now, for the volume here, since we're evolving about the x axis with the y equals, we will use the disc method. Which is V Equals pi multiplied by the integral from a to b. Of F of X 2 d x Let's go ahead and set up our integral. We have v equals pi integral. From 0 to 8. Of 1 divided by X to the 2/3, all squared, DX. We can actually simplify this a little bit, because we have X 2/3 and the denominator. We can write this as pi integral from 0 to 8 of X-4/3 DX. Because we had the square term. Now, we notice here that this is actually an improper integral. This is because the function x -4/3 is not defined at 0. So we have to use an improper integral. We will say V equals pi. Multiplied by the limit as A approaches 0 from the right of the integral A to 8, x to the 4/3, DX. Now from here, we can solve our integral. V equals pi limit. A approaches zero from the right. Our antiderivative of X 4/3 is X-1/3 divided by -1/3 based on the power role. We have pi limit A approaches 0 from the right. Of -3 X to the -1/3, from A to 8. We can now plug in our limit and our bounce. We have pi, limit. A approaches 0 from the right. Of -3 multiplied by 8 to 1/3. -3 multiplied by a to -1/3. Now we notice here if we plug in 0. We actually end up getting infinity, because we would have 1 divided by 0. That limit goes to infinity. So we have pi multiplied by infinity. Which this is equals infinity meaning This volume diverges. We can then confirm then the answer to our problem is answer d. The volume integral diverges. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

90. Consider the infinite region in the first quadrant bounded by the graphs ofy = 1 / √x, y = 0, x = 0, and x = 1.b. Find the volume of the solid formed by revolving the region (i) about the x-axis

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90. Consider the infinite region in the first quadrant bounded by the graphs of
y = 1 / √x, y = 0, x = 0, and x = 1.
b. Find the volume of the solid formed by revolving the region (i) about the x-axis