Consider the region bounded by the graphs of
y = arctan(...

Question

Consider the region bounded by the graphs of

y = arctan(x), y = 0, and x = 1.

b. Find the volume of the solid formed by revolving this region about the y-axis.

Accepted Answer

Consider the region bounded by Y equals tangent inverse of 3 X. Y equals 0 and X equals 1/3. Compute the volume of the solid generated when this region is revolved about the y axis. Now, to solve this, we will make use of the shell method formula. This is given by V equals 2 pi integral from A to B. Of the radius multiplied by the height. D X Now, our radius is given by absolute value X minus C, where C is a line we're rotating around. But in this case, the Y axis means C is zero, so we just have X. For the height We have f of x minus g of x. Now because there's no 2nd equation, we just have y equals 0. We have Tent inverse 3 x as our height. We can now write our integral as 2 pi integral from 0 to 1/3. Of X tangent inverse 3 X. D X Now, we have an integration by parts. What we can do here then is use the formula, which is the integral UDV equals UV minus VDU with the integral. Now, in our case, U is going to be tangent inverse. 3 X. DU then will be the derivative, which is 1 divided by 1 + 3 X 2, or 1 divided by 1 + 9 X 2. This is also a chain rule, which means we have the inside derivative of 3, so we'll multiply this by 3. We have 3 divided by 1 plus 9 x 2. This will also have a DX at the very end. TV then Is given by X. Which means V is just the entire derivative of X2 divided by 2. We can now write are integral. 2 pi multiplied. By UV, which is X2 divided by 2. Tangent inverse 3 x. Minus the integral x 2 divided by 2 multiplied. By 3 divided by 1 plus 9 X 2 DX. Now we can simplify from here. We have 2 pi multiplied, x 2 divided by 2, tangent inverse 3 x. Midas 3 halves multiplied by the integral x2 divided by 1 + 9 X 2 DX. Now, for our fraction here, we can use algebraic manipulation and rewrite this. We can factor out a 1/9 and do some manipulation with x2 to get the following. We have 2 pi, X 2 divided by 2, tangent inverse, 3 X. Midas. Now we have our 3 halves. Integral Of 1/9 multiplied by 1 minus 1 divided by 1 + 9 X 2. Now we can evaluate our integral. Because we simplified by turning it into a subtraction, we can take the integral of both parts. I'm actually gonna factor out this 19. And put it with the 3 halves. So we have 2 pi. X 2 divided by 2 tangent inverse 3 x. -3 halves multiplied by 1/9. Multiplied again by the interval of 1, which is just x. Minus the integral of 1 divided by 1 + 9 x 2. That is a use substitution. That gives us 1/3. Tangent inverse of 3 x. And this is all from 0 to 1/3. Now we just need to plug in our bounce. When we plug in 1/3. We get 2 pi multiplied. Bay Pi minus 2 divided by 36. When we plug in 0, we get 0. So we then have. Toupee Multiplied by pi minus 2 divided by 36. Which then simplifies. We ain't up kidding Pi squad -2. Divided by 18. And this is in cubic units. This is the answer to our problem. I hope to help you solve it. Thank you for watching. Goodbye.

Consider the region bounded by the graphs ofy = arctan(x), y = 0, and x = 1.b. Find the volume of the solid formed by revolving this region about the y-axis.

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Consider the region bounded by the graphs of
y = arctan(x), y = 0, and x = 1.
b. Find the volume of the solid formed by revolving this region about the y-axis.