What is the integral from 0 to 3 of 1/(sqrt(36-z^2)) ?

The integral from 0 to 3 of 1/(sqrt(36-z^2)) is pi/6

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integral from 0 to 3 of 1/(sqrt(36-z^2))

\int_{0}^{3} \frac{1}{36 - z ^{2}} d z

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{0}^{3} \frac{1}{36 - z ^{2}} d z

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 1 d u

= \int_{0}^{\frac{π}{6}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{6}}

Simplify

= [u]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6}

= \frac{π}{6}

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\int_{- 4}^{2} (8 - 2 x - x^{2}) d x

What is the integral from 0 to 3 of 1/(sqrt(36-z^2)) ?
The integral from 0 to 3 of 1/(sqrt(36-z^2)) is pi/6