Find volume bounded by curve $x=\sqrt {25-y^2}$, and the line $x=3$, rotated about the $y$-axis

Question

My attempt:

$$2\pi\int_0^{5} (25-y^2) dy$$

What's my error?

Answer 1

Hint:

$x = 3$ and $x = \sqrt{25-y^2}$ intersect at $y = 4$ and $-4$

$$V = \pi \int_{-4}^{4} (25-y^2 - 9)dy = \pi \int_{-4}^{4} (16-y^2)dy$$

if you evaluate the integral you get $V = \frac{256\pi}{3}$

The Shell method of solving:

Radius $= x$ and height = $2\sqrt{25-x^2}$

$V = 2\pi \int_{3}^{5} 2x\sqrt{25-x^2}dx = \frac{256\pi}{3}$

Answer 2

You must subtract the area of the segment formed by the $x=3$. Try taking the integrals in sections:

• The integral of {the circle minus the line}
• Subtract that from the total integral of the circle