Answer to Question #196137 in Calculus for Angelo

Question

Find the volume of the solid generated when the region R bounded by the given curves is revolved about the indicated axis:

Circular Disk Method:

1. R:y=√𝑥, 𝑦=4, 𝑦−𝑎𝑥𝑖𝑠 axis of revolution: 𝑦=4
2. 𝑅:𝑦=𝑥^2, 𝑥=3, 𝑥−𝑎𝑥𝑖𝑠 axis of revolution: 𝑥=3

Circular Ring Method:

1. 𝑅: 𝑦=𝑥^2, 𝑦=2𝑥 axis of revolution: 𝑥−𝑎𝑥𝑖𝑠
2. 𝑅:𝑦^2=4𝑥, 𝑥^2=4𝑦 axis of revolution: 𝑥=4

Cylindrical Shell Method:

1. 𝑅:𝑦=𝑥^2, 𝑦=2𝑥 axis of revolution: 𝑥−𝑎𝑥𝑖𝑠
2. 𝑅: first-quadrant region bounded by 𝑦=4𝑥 and 𝑥^3=𝑦 axis of revolution: 𝑥=2
3. An oil tank in the shape of a sphere has a diameter of 60 ft. How much oil does the tank contain if the depth of the oil is 25ft?

Answer 1

1

$\pi\int\limits_0^16 (4-\sqrt{x})^2dx=\pi\int\limits_0^4 (16-8\sqrt{x}+x)dx=\\ \pi (16x-\frac{16}{3}(\sqrt{x})^3+\frac{x^2}{2})|_{x=0}^{x=16}dx=\pi(256-\frac{1024}{3}+\frac{256}{2})=\\ \frac{128}{3}\pi$

2

$\pi\int\limits_0^9 (3-\sqrt{y})^2dy=\pi\int\limits_0^9 (9-6\sqrt{y}+y)dy=\\ \pi (9y-\frac{12}{3}(\sqrt{y})^3+\frac{y^2}{2})|_{y=0}^{y=9}dx=\pi(81-108+\frac{81}{2})=\\ \frac{27}{2}\pi$

3

$\pi\int\limits_0^4(\sqrt{y})^2-(\frac{y}{2})^2dy=\pi(\frac{y^2}{2}-\frac{y^3}{12})|_{y=0}^{y=4}=\\ \pi(8-\frac{16}{3})=\frac{8}{3}\pi$

4

$\pi\int\limits_0^4(4-\frac{y^2}{2})^2-(4-2\sqrt{y})^2dy=\pi(16y-\frac{4y^3}{3}+\frac{y^5}{20}-16y+\frac{32}{3}(\sqrt{y})^3-2y^2)|_{y=0}^{y=4}=\pi (-\frac{256}{3}+51.2+\frac{256}{3}-32)=\\ =19.2\pi$

5

$\pi\int\limits_0^2x(2x-x^2)dx=\pi(\frac{16}{3}-\frac{16}{4})=\frac{4}{3}\pi$

6

$\frac{\pi}{2}\int\limits_0^2(2-x)(4x-x^3)dx=\frac{\pi}{2}(-\frac{32}{3}+\frac{32}{5}+16-8)=(4-\frac{32}{15})\pi=\frac{28}{15}\pi$

7

$\pi\int\limits_{0}^{35}x(\sqrt{60^2-x^2}-\sqrt{60^2-35^2})dx=\pi(\frac{-1}{3}((\sqrt{60^2-35^2})^3-60^3)-\frac{5}{2}\sqrt{95}x^2)=\\ =\pi(\frac{125}{3}(12^3-95\sqrt{95})-\frac{6125}{2}(\sqrt{95}))=\pi(72000-\frac{42125}{6}\sqrt{95})\approx 11214 ft^3$