What is area x^2+y^2=64,-8,3 ?

The answer to area x^2+y^2=64,-8,3 is 64(arcsin(3/8)+(pi+sin(2arcsin(3/8)))/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area x^2+y^2=64,-8,3

area

x^{2} + y^{2} = 64, - 8, 3

64 (arcsin (\frac{3}{8}) + \frac{π + sin ( 2 arcsin ( \frac{3}{8} ) )}{2})

Decimal

147.38095 \dots

Solution steps

Isolate

y

for

x^{2} + y^{2} = 64 : y = 64 - x^{2}, y = - 64 - x^{2}

Apply the area formula:

\int_{- 8}^{3} 2 64 - x^{2} d x

Solve

\int_{- 8}^{3} 2 64 - x^{2} d x : 64 (arcsin (\frac{3}{8}) + \frac{π + sin ( 2 arcsin ( \frac{3}{8} ) )}{2})

The area is:

= 64 (arcsin (\frac{3}{8}) + \frac{π + sin ( 2 arcsin ( \frac{3}{8} ) )}{2})

y^{^{''}} + 3 y = t^{3} - 1

\frac{d}{d x} (5 e^{x} - 3 x)

f (x) = (x - 9)^{2}

\int \frac{41 arctan ( x )}{x ^{2}} d x

f (x) = sin (ln (x))

What is area x^2+y^2=64,-8,3 ?
The answer to area x^2+y^2=64,-8,3 is 64(arcsin(3/8)+(pi+sin(2arcsin(3/8)))/2)