Answer:
The correct integral is [tex]V=\pi \int\limits^4_0 {[(\sqrt{25-y^{2}})^{2}-3^{2} }] \, dy[/tex] ⇒ 2nd answer
Step-by-step explanation:
The formula of the volume of a curve y = f(x) by integration around the y-axis is [tex]V=\pi \int\limits^b_a {[f(y)]^{2} } \, dy[/tex]
∵ 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 = [tex]\sqrt{25-y^{2}}[/tex]
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 = [tex]\sqrt{25-y^{2}}[/tex]
To find the volume integrate the difference of the squares of the equations of the two curves and multiply the difference by π
∵ [tex]V=\pi \int\limits^b_a {[[f(y)]^{2}-[g(y)]^{2}}] \, dy[/tex]
- Subtract square 3 from the square of [tex]\sqrt{25-y^{2}}[/tex]
∴ [tex]V=\pi \int\limits^4_0 {[(\sqrt{25-y^{2}})^{2}-3^{2} }] \, dy[/tex]
The correct integral is [tex]V=\pi \int\limits^4_0 {[(\sqrt{25-y^{2}})^{2}-3^{2} }] \, dy[/tex]
