What is the integral of ysqrt(x^2+y^2) ?

The integral of ysqrt(x^2+y^2) is 1/3 (x^2+y^2)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ysqrt(x^2+y^2)

\int y x^{2} + y^{2} d y

\frac{1}{3} (x^{2} + y^{2})^{\frac{3}{2}} + C

Solution steps

\int y x^{2} + y^{2} d y

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = x^{2} + y^{2}

= \frac{( x ^{2} + y ^{2} ) ^{3}}{3}

Simplify

\frac{( x ^{2} + y ^{2} ) ^{3}}{3} : \frac{1}{3} (x^{2} + y^{2})^{\frac{3}{2}}

= \frac{1}{3} (x^{2} + y^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (x^{2} + y^{2})^{\frac{3}{2}} + C

\int cos (x) 2^{s i n (x)} d x

\int \frac{sin ( 2 t )}{sin ( t )} d t

\int 2 cos (x) - 9 sin (x) d x

\int \frac{4}{x 16 - ln ( x ^{2} )} d x

\int 3 x^{\frac{4}{3}} d x

What is the integral of ysqrt(x^2+y^2) ?
The integral of ysqrt(x^2+y^2) is 1/3 (x^2+y^2)^{3/2}+C