Math / Calculus

Question $\iint_{0}\left(x^{2}+y^{3}\right) d x d y . \quad 0=\{(x, y) \mid 0 \leq x \leq 10 \leq y \leq 1\}$

Studdy Solution

STEP 1

What is this asking? We need to calculate the volume under the surface $z = x^2 + y^3$ over the square region where both $x$ and $y$ are between $\bf{0}$ and $\bf{1}$ . Watch out! Don't forget to apply the correct order of integration and the limits for each variable!

STEP 2

1. Set up the double integral
2. Integrate with respect to $x$
3. Integrate with respect to $y$

STEP 3

Alright, let's set up our double integral!
We're given the function $f(x, y) = x^2 + y^3$ and the region of integration $0 \leq x \leq 1$ and $0 \leq y \leq 1$ .
So, our double integral looks like this:
$\int_{0}^{1} \int_{0}^{1} (x^2 + y^3) \, dx \, dy$

STEP 4

Now, let's integrate with respect to $x$ first, treating $y$ as a constant.
Remember the power rule: $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ .
$\int_{0}^{1} \left[ \frac{x^3}{3} + xy^3 \right]_{0}^{1} dy$

STEP 5

Let's plug in our limits of integration for $x$ , which are $\bf{0}$ and $\bf{1}$ .
$\int_{0}^{1} \left( \left( \frac{1^3}{3} + 1 \cdot y^3 \right) - \left( \frac{0^3}{3} + 0 \cdot y^3 \right) \right) dy$

STEP 6

This simplifies to:
$\int_{0}^{1} \left( \frac{1}{3} + y^3 \right) dy$

STEP 7

Now, we integrate with respect to $y$ .
$\left[ \frac{1}{3}y + \frac{y^4}{4} \right]_{0}^{1}$

STEP 8

Let's plug in our limits of integration for $y$ , which are $\bf{0}$ and $\bf{1}$ .
$\left( \frac{1}{3}(1) + \frac{1^4}{4} \right) - \left( \frac{1}{3}(0) + \frac{0^4}{4} \right)$

STEP 9

This simplifies to:
$\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$

STEP 10

Our final answer is $\frac{7}{12}$ !

Was this helpful?

Question $\iint_{0}\left(x^{2}+y^{3}\right) d x d y . \quad 0=\{(x, y) \mid 0 \leq x \leq 10 \leq y \leq 1\}$

Studdy Solution

STEP 1

What is this asking? We need to calculate the volume under the surface $z = x^2 + y^3$ over the square region where both $x$ and $y$ are between $\bf{0}$ and $\bf{1}$ . Watch out! Don't forget to apply the correct order of integration and the limits for each variable!

STEP 2

1. Set up the double integral
2. Integrate with respect to $x$
3. Integrate with respect to $y$

STEP 3

Alright, let's set up our double integral!
We're given the function $f(x, y) = x^2 + y^3$ and the region of integration $0 \leq x \leq 1$ and $0 \leq y \leq 1$ .
So, our double integral looks like this:
$\int_{0}^{1} \int_{0}^{1} (x^2 + y^3) \, dx \, dy$

STEP 4

Now, let's integrate with respect to $x$ first, treating $y$ as a constant.
Remember the power rule: $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ .
$\int_{0}^{1} \left[ \frac{x^3}{3} + xy^3 \right]_{0}^{1} dy$

STEP 5

Let's plug in our limits of integration for $x$ , which are $\bf{0}$ and $\bf{1}$ .
$\int_{0}^{1} \left( \left( \frac{1^3}{3} + 1 \cdot y^3 \right) - \left( \frac{0^3}{3} + 0 \cdot y^3 \right) \right) dy$

STEP 6

This simplifies to:
$\int_{0}^{1} \left( \frac{1}{3} + y^3 \right) dy$

STEP 7

Now, we integrate with respect to $y$ .
$\left[ \frac{1}{3}y + \frac{y^4}{4} \right]_{0}^{1}$

STEP 8

Let's plug in our limits of integration for $y$ , which are $\bf{0}$ and $\bf{1}$ .
$\left( \frac{1}{3}(1) + \frac{1^4}{4} \right) - \left( \frac{1}{3}(0) + \frac{0^4}{4} \right)$

STEP 9

This simplifies to:
$\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$

STEP 10

Our final answer is $\frac{7}{12}$ !

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question∬0(x2+y3)dxdy.0={(x,y)∣0≤x≤10≤y≤1}\iint_{0}\left(x^{2}+y^{3}\right) d x d y . \quad 0=\{(x, y) \mid 0 \leq x \leq 10 \leq y \leq 1\}∬0​(x2+y3)dxdy.0={(x,y)∣0≤x≤10≤y≤1}

Studdy Solution

STEP 1

What is this asking? We need to calculate the volume under the surface z=x2+y3z = x^2 + y^3z=x2+y3 over the square region where both xxx and yyy are between 0\bf{0}0 and 1\bf{1}1. Watch out! Don't forget to apply the correct order of integration and the limits for each variable!

STEP 2

1. Set up the double integral2. Integrate with respect to xxx3. Integrate with respect to yyy

STEP 3

STEP 4

Now, let's **integrate** with respect to xxx first, treating yyy as a constant.Remember the power rule: ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}∫xndx=n+1xn+1​.∫01[x33+xy3]01dy\int_{0}^{1} \left[ \frac{x^3}{3} + xy^3 \right]_{0}^{1} dy∫01​[3x3​+xy3]01​dy

STEP 5

Let's **plug in** our limits of integration for xxx, which are 0\bf{0}0 and 1\bf{1}1.∫01((133+1⋅y3)−(033+0⋅y3))dy\int_{0}^{1} \left( \left( \frac{1^3}{3} + 1 \cdot y^3 \right) - \left( \frac{0^3}{3} + 0 \cdot y^3 \right) \right) dy∫01​((313​+1⋅y3)−(303​+0⋅y3))dy

STEP 6

This **simplifies** to:∫01(13+y3)dy\int_{0}^{1} \left( \frac{1}{3} + y^3 \right) dy∫01​(31​+y3)dy

STEP 7

Now, we **integrate** with respect to yyy.[13y+y44]01\left[ \frac{1}{3}y + \frac{y^4}{4} \right]_{0}^{1}[31​y+4y4​]01​

STEP 8

Let's **plug in** our limits of integration for yyy, which are 0\bf{0}0 and 1\bf{1}1.(13(1)+144)−(13(0)+044)\left( \frac{1}{3}(1) + \frac{1^4}{4} \right) - \left( \frac{1}{3}(0) + \frac{0^4}{4} \right)(31​(1)+414​)−(31​(0)+404​)

STEP 9

This **simplifies** to:13+14=412+312=712\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}31​+41​=124​+123​=127​

STEP 10

Our **final answer** is 712\frac{7}{12}127​!

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question $\iint_{0}\left(x^{2}+y^{3}\right) d x d y . \quad 0=\{(x, y) \mid 0 \leq x \leq 10 \leq y \leq 1\}$

What is this asking? We need to calculate the volume under the surface $z = x^2 + y^3$ over the square region where both $x$ and $y$ are between $\bf{0}$ and $\bf{1}$ . Watch out! Don't forget to apply the correct order of integration and the limits for each variable!

1. Set up the double integral
2. Integrate with respect to $x$
3. Integrate with respect to $y$

Now, let's integrate with respect to $x$ first, treating $y$ as a constant.
Remember the power rule: $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ .
$\int_{0}^{1} \left[ \frac{x^3}{3} + xy^3 \right]_{0}^{1} dy$

Let's plug in our limits of integration for $x$ , which are $\bf{0}$ and $\bf{1}$ .
$\int_{0}^{1} \left( \left( \frac{1^3}{3} + 1 \cdot y^3 \right) - \left( \frac{0^3}{3} + 0 \cdot y^3 \right) \right) dy$

This simplifies to:
$\int_{0}^{1} \left( \frac{1}{3} + y^3 \right) dy$

Now, we integrate with respect to $y$ .
$\left[ \frac{1}{3}y + \frac{y^4}{4} \right]_{0}^{1}$

Let's plug in our limits of integration for $y$ , which are $\bf{0}$ and $\bf{1}$ .
$\left( \frac{1}{3}(1) + \frac{1^4}{4} \right) - \left( \frac{1}{3}(0) + \frac{0^4}{4} \right)$

This simplifies to:
$\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$

Our final answer is $\frac{7}{12}$ !