For problem 2, you are correct in stating that a curve forms. Specifically, if we were to trace along the outer edge of the shape, then we'd form a <u>parabola</u> that has been tilted 45 degrees compared to the more familiar form that students are taught (where the axis of symmetry is vertical).

For more information, search out "Tangent method for parabolas". As the name implies, the tangent method draws out the tangents of the parabola which helps form the parabola itself.

Everything else on your paper is correct. You have problem 1 correct, and the table is filled out perfectly. Nice work.

4 0

3 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

2 years ago

Find the difference between the longest and shortest bean sprouts <br><br>Please i need

aniked [119]

Answer:

7/4 inches

Step-by-step explanation:

Longest bean sprout = 8/8 + 8/8

Longest bean sprout = 1 + 1

Longest bean sprout = 2

Shortest bean sprout = 2/8

Difference = 2 - 2/8

Difference = (16-2)/8

Difference = 14/8

Difference = 7/4

Hence the distance is 7/4 inches

5 0

3 years ago

Can someone help please

vova2212 [387]

Answer:

24sq.ft

Step-by-step explanation:

<h3>Area of a parallelogram = Base*Height</h3>

Here,

→Base = 6ft, Height = 4ft

→ 6*4ft

→24sq.ft

<h3 /><h3>While writing area of a figure, we have to express the area in sq which means square</h3>

<h2> <em>ThankYou</em></h2>

<em>Please mark me as brainliest</em>

7 0

3 years ago