Question

Which of the following integrals correctly computes the volume formed when the region bounded by the curves x2 + y2 = 25, x = 3, and y = 0 is rotated around the y-axis?

Answer 1

The correct integral is
`V=\pi \int\limits^4_0 {[(\sqrt{25-y^(2)})^(2)-3^(2) }] \, dy` ⇒ 2nd answer

Explanation:

The formula of the volume of a curve y = f(x) by integration around the y-axis is
`V=\pi \int\limits^b_a {[f(y)]^(2) } \, dy`

∵ The curves are x² + y² = 25 , x = 3 and y = 0

- Lets find x in terms of y in the first curve

∵ x² + y² = 25

- Subtract y² from both sides

∴ x² = 25 - y²

- Take square root for both sides

∴ x =
`\sqrt{25-y^(2)}`

To find the limit of the integration b substitute x by 3

∵ x = 3

∵ x² + y² = 25

∴ 3² + y² = 25

∴ 9 + y² = 25

- Subtract 9 from both sides

∴ y² = 16

- Take square root for both sides

∴ y = ± 4

∵ The 1st limit a is at y = 0

∴ a = 0

∵ The 2nd limit is b at y = 4

∴ b = 4

∴ b = 4

There are two curves x = 3 and x =
`\sqrt{25-y^(2)}`

To find the volume integrate the difference of the squares of the equations of the two curves and multiply the difference by π

∵
`V=\pi \int\limits^b_a {[[f(y)]^(2)-[g(y)]^(2)}] \, dy`

- Subtract square 3 from the square of
`\sqrt{25-y^(2)}`

∴
`V=\pi \int\limits^4_0 {[(\sqrt{25-y^(2)})^(2)-3^(2) }] \, dy`

The correct integral is
`V=\pi \int\limits^4_0 {[(\sqrt{25-y^(2)})^(2)-3^(2) }] \, dy`