1
$\begingroup$

I cannot figure out how to solve this. The region is bounded by $xy=7$, $y=1$, $y=8$ and $x=9$. Revolving about the line $x=9$.

I came up with the following integral to solve this:

$\pi\int_{1}^{8}(9)^2-(7/y)^2dy$

The answer I got is $1646.587$.

What am I doing wrong? Did I set up the problem incorrectly?

enter image description here

  • 0
    Could it be that you take the difference of the squares instead of the square of the difference? – 2017-01-27
  • 0
    I have tried $\pi\int_{1}^{8}(9-(7/y))^2dy$ as well, but it still says my answer is incorrect? Doing it this way I got 1092.851. – 2017-01-27
  • 0
    So, what is the "correct" answer according to your source? – 2017-01-27
  • 0
    I do not know the correct answer. It is for a Calculus 2 WebAssign homework assignment. – 2017-01-27
  • 0
    I dont know that tool, but maybe you have to enter the exact answer, with pi and logarithms? – 2017-01-27
  • 0
    Yea I am not sure but I have been trying to figure this one problem out for a while now. Everything looks right doesn't it? – 2017-01-27
  • 0
    Hit the "Talk to tutor" button? :) – 2017-01-27

1 Answers 1

0

A general tip in the case you are not revolving around a coordinate axis is to transform your expressions so that you are. Your situation looks like this:

graphs1

If you shift $9$ steps to the left (i.e. replace all $x$ by $x+9$), then it looks like this

graphs2

So, the domain that revolves is bounded by the curves $$ (x+9)y=7,\quad y=1,\quad y=8,\quad\text{and}\quad x=0. $$ The first equation gives $$ x=\frac{7}{y}-9. $$ Thus, the volume you look for is given by the integral $$ \int_1^8\pi(7/y-9)^2\,dy. $$