Answer to Question #144710 in Calculus for Besmallah Yousefi

Question #144710
2. a) Calculate the volume under the plane z=x+2y and over the region R={(x,y)|1≤x≤2 and 3≤y≤5}.
b) Calculate the volume under the surface z=xy^2+y^3 and the region R={(x,y)|0≤x≤2 and 1≤y≤3},
1
Expert's answer
2020-11-17T16:58:42-0500

If z=f(x,y) - surface, then volume under this surface is calculated as Rf(x,y)dxdy\int_R f(x,y)dxdy where R is a region of xOy plane.


2.a) 12dx35(x+2y)dy=12(xy+y2)35dx=\int_1^2dx\int_3^5(x+2y)dy=\int_1^2(xy+y^2)|_3^5 dx=


=12(5x+253x9)dx=12(2x+16)dx==\int_1^2(5x+25-3x-9) dx=\int_1^2(2x+16) dx=


=(x2+16x)12=4+32116=19=(x^2+16x)|_1^2=4+32-1-16=19 .


Answer: 19


2.b) 02dx13(xy2+y3)dy=02(xy33+y44)13dx=\int_0^2dx\int_1^3(xy^2+y^3)dy=\int_0^2(\frac{xy^3}{3}+\frac{y^4}{4})|_1^3dx=


=02(9x+814x314)dx=02(26x3+20)dx==\int_0^2(9x+\frac{81}{4}-\frac{x}{3}-\frac{1}{4})dx=\int_0^2(\frac{26x}{3}+20)dx=


=(13x23+20x)02=523+40=1723=5713=(\frac{13x^2}{3}+20x)|_0^2=\frac{52}{3}+40=\frac{172}{3}=57\frac{1}{3} .


Answer: 571357\frac{1}{3}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment