Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Question

Answered step-by-step

Sketch the region enclosed by the given curves and find its area.

$ y = \sec^2 x $ , $ y = 8 \cos x $ , $ \frac{-\pi}{3} \le x \le \frac{\pi}{3} $

Video Answer

Solved by verified expert

This problem has been solved!

Try Numerade free for 7 days

Like

Report

Official textbook answer

Video by Sky Li

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

Missouri State University

Harvey Mudd College

Boston College

Lectures

03:00

Sketch the region enclosed…

03:41

Find the area of the regio…

04:09

Find the areas of the regi…

03:12

03:36

00:43

04:25

05:34

18:46

04:54

02:27

02:41

On this problem, we need to sketch the reeds been enclosed by those 2 curves. So let's draw this x y plane first and as we can see, the second 1 is much easier to draw and we know our x. Our domain goes from negative pi over 3 to pi over 3, so assuming this is pi over 3. This is negative pi over 3 and here is positive, pi over 3 or cosine x. The goes to 0 point. We know this is a given function. Sorry, symmetric along y exists, so x, equal to 0 is equals to 8 and we know that x equals to pi over 3 cosine pi over 3 is half so this equals to 4 since its symmetric. This also equals 4 point. So y equals to 8 cosine x should look like this okay. This is some y equals to cosine, 8 cosine x and the second. The first term here is second x square, and we know the second x square is just over. Cosine x square is 1 over cold, fine, x square, and so the x equals to 0. This equals 1 and x equals 2 pi over 3. On this cosine power, 3 is put half so 1 half square is 1 quarter and the reciprocal of our quarter is 4. Actually, intersection point is exactly of the ends of cosine x to the end poinsnegative pi over 3 and a positive pi over 3. These functions should look like this. We got this enclosed area now. Our next step is to find this area. So, as we see, all the things represented by x will take the integral with respect to x and the boundary we already found is just a negative pi over 3 to pi over 3 and the thing inside the integral is upper curve. Upper curve is 8 cos. Minus the lower curve, which is a second x square now, let's find the antidotive of this, so the first term will be 8 sine x. They know the antidote of cosine x is just 6. If you remember, anti derivative for second x is actually tangent x. If you don't remember, you can check the charge of the anti derivative, then we evaluate the boundary x equals to pi over 3 and x, equals to negative pi over 3, and then our x equals to pi over 3. We know sine pi over 3 square root. 3 over 2 point, so these 4 square root, 3 minus and tangent tangent pi over 3 square root, 3 minus the plug and pi x equals to the negative power 3. So this will be negative. 4 square 3 plus square root 3 right. So our answer will be 6 square root. Thre.

View More Answers From This Book

Find Another Textbook

01:08

Given M1 = 25, M2 = 28, N = 5, and sD = 7.00, calculate t.

03:04

"Hot Tamales" are chewy, cinna mon-flavored candies. A bulk vendin…

00:59

The distributions of X and of Y are described here. If X and Y are independe…

01:04

In problem, find the domain of each function.f(x) = x/x2 + 1

01:12

In problem, find the domain of each function.f(x) = x2/x2 + 1

02:20

In problem, find the domain of each function.g(x) = x/x2 - 16

00:45

Dawn decides to invest $2000 per year in stock at the beginning of each of t…

01:03

Two jurors are selected from 4 alternates to serve at a murder trial. Using …

03:05

In a random sample of 12 mobile devices, the mean repair cost was $90.42 and…

08:12

Approximately 60% of Americans have some form of life insurance coverage. Fo…