Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 5 sin(x), y = 5 cos(x), 0 ≤ x ≤ π/4; about y = −1

Answer 1

Step-by-step explanation:

First we need to find which function is the outside radius and which is the inside radius (by which I mean which is farther and which is closer to the axis of rotation).  We can do this by graphing, but we can also do this by evaluating each function at the end limits.

At x = 0:

y = 5 sin 0 = 0

y = 5 cos 0 = 5

At x = π/4:

y = 5 sin π/4 = 5√2 / 2

y = 5 cos π/4 = 5√2 / 2

So y = 5 cos x is the outside radius because it is farther away from y = -1, and y = 5 sin x is the inside radius because it is closer to y = -1.

Here's a graph:

desmos.com/calculator/5oaiobcpww

The volume of the rotation is:

V = π ∫ₐᵇ [(R−y)² − (r−y)²] dx

where a is the lower limit, b is the upper limit, R is the outside radius, r is the inside radius, and y is the axis of rotation.

Plugging in:

V = π ∫₀ᵖ [(5 cos x − -1)² − (5 sin x − -1)²] dx

V = π ∫₀ᵖ [(5 cos x + 1)² − (5 sin x + 1)²] dx

V = π ∫₀ᵖ [(25 cos² x + 10 cos x + 1) − (25 sin² x + 10 sin x + 1)] dx

V = π ∫₀ᵖ [25 cos² x + 10 cos x + 1 − 25 sin² x − 10 sin x − 1] dx

V = π ∫₀ᵖ [25 (cos² x − sin² x) + 10 cos x − 10 sin x] dx

V = π ∫₀ᵖ [25 cos (2x) + 10 cos x − 10 sin x] dx

V = π [25/2 sin (2x) + 10 sin x + 10 cos x] from 0 to π/4

V = π [25/2 sin (π/2) + 10 sin (π/4) + 10 cos (π/4)] − π [25/2 sin 0 + 10 sin 0 + 10 cos 0]

V = π [25/2 + 5√2 + 5√2] − 10π

V = π (5/2 + 10√2)

V ≈ 52.283

Answer 2

In this exercise it is necessary to calculate the volume of the rotating solid, in this way we have:

$V= 52.3$

First we need to find which function is the outside radius and which is the inside radius (by which I mean which is farther and which is closer to the axis of rotation).  We can do this by graphing, but we can also do this by evaluating each function at the end limits.

$x=0\\y=5sin(x)\\y=5sin(0)=0\\y=5cos(0)= 5\\\\x=\pi/4\\y=5sin(\pi/4)= 5(\sqrt{2/2})\\y=5cos( \pi/4)= 5/\sqrt{2}$

So $y = 5 cos x$ is the outside radius because it is farther away from y = -1, and $y = 5 sin x$ is the inside radius because it is closer to y = -1. Here's a graph (first image). The volume of the rotation is:

$V=\pi\int\limits^a_b {[(R-y)^2-(r-y)^2]} \, dx$

Where a is the lower limit, b is the upper limit, R is the outside radius, r is the inside radius, and y is the axis of rotation. Plugging in:

$V=\pi\int\limits^a_b {[(R-y)^2-(r-y)^2]} \, dx\\= \pi\int\limits^p_0 {[(5 cos(x) + 1)^2-(5 sin(x) + 1)^2]} \, dx\\=\pi\int\limits^p_0 {[(25 cos^2 (x) + 10 cos (x) + 1)-(25 sin^2 (x) + 10 sin (x) + 1)]} \, dx\\=\pi\int\limits^p_0 {[(5 cos^2 (x) + 10 cos (x) + 1 -25 sin^2 (x)- 10 sin (x)- 1]} \, dx\\=\pi\int\limits^p_0 {[(25 (cos^2( x)- sin^2( x)) + 10 cos( x)- 10 sin( x)]} \, dx\\=\pi\int\limits^p_0 {[(25 cos (2x) + 10 cos (x)- 10 sin (x)]} \, dx$

$=\pi[(25/2 sin (2x) + 10 sin x + 10 cos (x)]\\\\=\pi[(25/2 sin (\pi/2) + 10 sin (\pi/4) + 10 cos (\pi/4)] - \pi [25/2 sin( 0) + 10 sin( 0) + 10 cos(0)]\\\\=\pi[25/2+5\sqrt{2}+5/\sqrt{2}]-10\pi\\\\\V= 52.3$

See more about volume of the solid at brainly.com/question/10956718