What is the integral from-t to t of sqrt(1-x^2-y^2) ?

The integral from-t to t of sqrt(1-x^2-y^2) is 0

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integral from-t to t of sqrt(1-x^2-y^2)

\int_{- t}^{t} 1 - x^{2} - y^{2} d y

0

Solution steps

\int_{- t}^{t} 1 - x^{2} - y^{2} d y

Apply Trigonometric Substitution

: \int_{a r c s i n (- \frac{t}{1 - x ^{2}})}^{a r c s i n (\frac{t}{1 - x ^{2}})} cos (u) - x^{2} + x^{2} sin^{2} (u) + cos^{2} (u) 1 - x^{2} d u

= \int_{a r c s i n (- \frac{t}{1 - x ^{2}})}^{a r c s i n (\frac{t}{1 - x ^{2}})} cos (u) - x^{2} + x^{2} sin^{2} (u) + cos^{2} (u) 1 - x^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 1 - x^{2} \cdot \int_{a r c s i n (- \frac{t}{1 - x ^{2}})}^{a r c s i n (\frac{t}{1 - x ^{2}})} cos (u) - x^{2} + x^{2} sin^{2} (u) + cos^{2} (u) d u

Apply u-substitution

: \int_{c o s (a r c s i n (- \frac{t}{1 - x ^{2}}))}^{c o s (a r c s i n (\frac{t}{1 - x ^{2}}))} - \frac{v ^{2} - x ^{2} + 1}{1 - v ^{2}} d v

= 1 - x^{2} \cdot \int_{c o s (a r c s i n (- \frac{t}{1 - x ^{2}}))}^{c o s (a r c s i n (\frac{t}{1 - x ^{2}}))} - \frac{v ^{2} - x ^{2} + 1}{1 - v ^{2}} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 1 - x^{2} - - x^{2} + 1 \cdot \int_{c o s (a r c s i n (- \frac{t}{1 - x ^{2}}))}^{c o s (a r c s i n (\frac{t}{1 - x ^{2}}))} \frac{v ^{2}}{1 - v ^{2}} d v

Apply Trigonometric Substitution

: \int_{a r c s i n (c o s (a r c s i n (- \frac{t}{1 - x ^{2}})))}^{a r c s i n (c o s (a r c s i n (\frac{t}{1 - x ^{2}})))} sin^{2} (w) d w

= 1 - x^{2} - - x^{2} + 1 \cdot \int_{a r c s i n (c o s (a r c s i n (- \frac{t}{1 - x ^{2}})))}^{a r c s i n (c o s (a r c s i n (\frac{t}{1 - x ^{2}})))} sin^{2} (w) d w

Rewrite using trig identities

= 1 - x^{2} - - x^{2} + 1 \cdot \int_{a r c s i n (c o s (a r c s i n (- \frac{t}{1 - x ^{2}})))}^{a r c s i n (c o s (a r c s i n (\frac{t}{1 - x ^{2}})))} \frac{1 - cos ( 2 w )}{2} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 1 - x^{2} - - x^{2} + 1 \frac{1}{2} \cdot \int_{a r c s i n (c o s (a r c s i n (- \frac{t}{1 - x ^{2}})))}^{a r c s i n (c o s (a r c s i n (\frac{t}{1 - x ^{2}})))} 1 - cos (2 w) d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 1 - x^{2} - - x^{2} + 1 \frac{1}{2} \int_{a r c s i n (c o s (a r c s i n (- \frac{t}{1 - x ^{2}})))}^{a r c s i n (c o s (a r c s i n (\frac{t}{1 - x ^{2}})))} 1 d w - \int_{a r c s i n (c o s (a r c s i n (- \frac{t}{1 - x ^{2}})))}^{a r c s i n (c o s (a r c s i n (\frac{t}{1 - x ^{2}})))} cos (2 w) d w

\int_{a r c s i n (c o s (a r c s i n (- \frac{t}{1 - x ^{2}})))}^{a r c s i n (c o s (a r c s i n (\frac{t}{1 - x ^{2}})))} 1 d w = 0

\int_{a r c s i n (c o s (a r c s i n (- \frac{t}{1 - x ^{2}})))}^{a r c s i n (c o s (a r c s i n (\frac{t}{1 - x ^{2}})))} cos (2 w) d w = 0

= 1 - x^{2} (- - x^{2} + 1 \frac{1}{2} (0 - 0))

Simplify

1 - x^{2} (- - x^{2} + 1 \frac{1}{2} (0 - 0)) : 0

= 0

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What is the integral from-t to t of sqrt(1-x^2-y^2) ?
The integral from-t to t of sqrt(1-x^2-y^2) is 0